devbox
羊村懒王
这个屌丝很懒，什么也没留下！
关注作者
热门标签
热门文章
当前位置:   article > 正文

2023西工大NOJ (C语言版) 完结！！！

作者：羊村懒王 | 2024-03-20 07:48:18

赞

踩

2023西工大noj

前言

代码已同步至 gitee：2023NOJ（C语言版）！

题目过于垃圾和睿智，做这种题简直是浪费生命，甚至会让代码水平剧烈下降，也就81-90题值得做一下，有这功夫不如多打会儿游戏。
贴出 100 题中全部 AC 代码，并不保证代码正确性、可读性、简洁性、最佳性，只能保证AC；目前 4 道题 WA，还有 5 道题使用 CPP。具体情况如下：

  • 41-50：飞机起飞速度（WA）；
  • 51-60：字符串替换（WA）；
  • 61-70：GPS通信协议（CPP）；
  • 71-80：卫星定位（WA），日出日落时间（WA），火箭发射模拟（CPP），晶体结构（CPP），原子计数（CPP）；
  • 81-90：危险的组合（CPP）。

考试模板

作者考试时会携带的资料如下：

  1. // 字符串操作函数(常用)
  2. // 为避免内存泄漏和野指针问题, 通常只能对开在栈上字符串使用, 而不能对开在堆上的字符串使用
  3. # include <string.h>
  4. // 失败时均返回NULL
  5. void *memset(void *str, int c, size_t n); // 复制字符c到str的前n个字符.
  6. // 使用memset对数组初始化时, 字符c只能为0或-1或0x3f.
  7. void *memcpy(void *dest, const void *src, size_t n); // 从src复制n个字符到dest, 一般不用于src和dest内存重叠的情况.
  8. void *memmove(void *dest, const void *src, size_t n); // 从src复制n个字符到dest, src和dest内存重叠时推荐使用.
  9. size_t strlen(const char *str); // 返回str长度, 即头指针到'\0'的距离(但不包含'\0').
  10. char *strcat(char *dest, const char *src); // 将src追加到dest结尾.
  11. char *strchr(const char *str, int c); // 返回str中第一次出现字符c的位置.
  12. char *strstr(const char *haystack, const char *needle); // 返回在haystack中第一次出现needle的起始位置.
  13. size_t strspn(const char *str1, const char *str2); // str1中第一个不在str2出现的字符的下标
  14. char *strtok(char *str, const char *delim); // 以delim为分隔, 分解str.
  1. // 字符串操作函数(常用)
  2. // 为避免内存泄漏和野指针问题, 通常只能对开在栈上字符串使用, 而不能对开在堆上的字符串使用
  3. # include <string.h>
  4. // 失败时均返回NULL
  5. void *memset(void *str, int c, size_t n); // 复制字符c到str的前n个字符.
  6. // 使用memset对数组初始化时, 字符c只能为0或-1或0x3f.
  7. void *memcpy(void *dest, const void *src, size_t n); // 从src复制n个字符到dest, 一般不用于src和dest内存重叠的情况.
  8. void *memmove(void *dest, const void *src, size_t n); // 从src复制n个字符到dest, src和dest内存重叠时推荐使用.
  9. size_t strlen(const char *str); // 返回str长度, 即头指针到'\0'的距离(但不包含'\0').
  10. char *strcat(char *dest, const char *src); // 将src追加到dest结尾.
  11. char *strchr(const char *str, int c); // 返回str中第一次出现字符c的位置.
  12. char *strstr(const char *haystack, const char *needle); // 返回在haystack中第一次出现needle的起始位置.
  13. size_t strspn(const char *str1, const char *str2); // str1中第一个不在str2出现的字符的下标
  14. char *strtok(char *str, const char *delim); // 以delim为分隔, 分解str.
  1. // 指针数组模板
  2. #include <stdio.h>
  3. #include <stdlib.h>
  4.  
  5. typedef struct {
  6.     char id[64];
  7.     int a;
  8.     int b;
  9. } PointerArray;
  10.  
  11. int main(){
  12.     int n; scanf("%d", &n);
  13.     PointerArray *arr[n];
  14.  
  15.     for (int i = 0; i < n; ++i) {
  16.         arr[i] = (PointerArray*) calloc (sizeof(PointerArray), 1);
  17.         scanf("%s %d %d", arr[i]->id, &arr[i]->a, &arr[i]->b);
  18.     }
  19.  
  20.     for (int i = 0; i < n; ++i)
  21.         printf("%s %d %d", arr[i]->id, arr[i]->a, arr[i]->b);
  22.     return 0;
  23. }
  1. // 二维数组模板
  2. #include <stdio.h>
  3. #include <stdlib.h>
  4.  
  5. int** init(int n) {
  6.     int** matrix = (int**)malloc(sizeof(int*) * (n + 1));
  7.     for(int i = 0; i <= n; i++) {
  8.         matrix[i] = (int*)malloc(sizeof(int) * (n + 1));
  9.     }
  10.     return matrix;
  11. }
  12.  
  13. int op(int **matrix, int n);
  14.  
  15. int main() {
  16.     int n;
  17.     scanf("%d", &n);
  18.     int **matrix = init(n);
  19.     for (int i = 1; i <= n; ++i) {
  20.         for (int j = 1; j <= n; ++j) {
  21.             scanf("%d", &matrix[i][j]);
  22.         }
  23.     }
  24.     printf("%d", op(matrix, n));
  25.     return 0;
  26. }
  1. // 快速排序模板: 从小到大
  2. void quickSort(int arr[], int left, int right) {
  3. 	if (left >= right) return;
  4. 	int flag = arr[(left + right) / 2];
  5. 	int head = left - 1, tail = right + 1;
  6. 	while (head < tail) {
  7. 		do head++; while (arr[head] < flag); // 若从大到小需修改本行
  8. 		do tail--; while (arr[tail] > flag); // 若从大到小需修改本行
  9. 		if (head < tail) {
  10. 			int tmp = arr[head];
  11. 			arr[head] = arr[tail], arr[tail] = tmp;
  12. 		}
  13. 	}
  14. 	quickSort(arr, left, tail);
  15. 	quickSort(arr, tail + 1, right);
  16. }
  1. // 冒泡排序模板: 从小到大
  2. #include <stdbool.h>
  3.  
  4. void bubble(int arr[], int n) {
  5.     for (int i = 0; i < n - 1; ++i) {
  6.         bool flag = false;
  7.         for (int j = 0; j < n - i - 1; ++j) {
  8.             if (arr[j] > arr[j + 1]) { // 若从大到小需修改本行
  9.                 int tmp = arr[j];
  10.                 arr[j] = arr[j + 1], arr[j + 1] = tmp, flag = true;
  11.             }
  12.         }
  13.         if (!flag) break;
  14.     }
  15. }
  1. // 二分查找模板
  2. long long left = 1, right = 1e19, mid, ans;
  3.  
  4. while(left <= right) {
  5.     mid = ((right - left) >> 1) + left;
  6.     if (isLeft(mid)) right = mid - 1, ans = mid;
  7.     else left = mid + 1;
  8. }
  1. // 文件读写模板
  2. #include <stdio.h>
  3. #include <stdlib.h>
  4.  
  5. int main(){
  6.     int n; scanf("%d",&n);
  7.  
  8.     FILE *file = fopen("rr.dat", "w");
  9.     for(int i=1;i<=n;i++)
  10.         fprintf(file, "%d\n", i);
  11.     fclose(file);
  12.  
  13.     file = fopen("rr.dat", "r");
  14.     while (fscanf(file, "%d", &ans) == 1)
  15.         printf("%d ", ans);
  16.     fclose(file);
  17. }

1-10

Hello World

  1. #include<stdio.h>
  2.  
  3. int main(){
  4.     printf("Hello World");
  5.     return 0;
  6. }

A+B

  1. #include<stdio.h>
  2.  
  3. int main(){
  4.     
  5.     int a = 0, b = 0;
  6.     scanf("%d %d",&a,&b);
  7.     int c = a+b;
  8.     printf("%d",c);
  9.     
  10.     return 0;
  11. }

数据类型大小及范围

  1. #include <stdio.h>
  2. #include <limits.h>
  3.  
  4. int main(){
  5.     int id = 0;
  6.     scanf("%d",&id);
  7.     switch (id) {
  8.         case 1: // char
  9.             printf("%llu,%d,%d\n",sizeof(char),CHAR_MIN,CHAR_MAX);
  10.             break;
  11.         case 2: // unsigned char
  12.             printf("%llu,%u,%u\n",sizeof(unsigned char),0,UCHAR_MAX);
  13.             break;
  14.         case 3: // short
  15.             printf("%llu,%hd,%hd\n",sizeof(short),SHRT_MIN,SHRT_MAX);
  16.             break;
  17.         case 4: // unsigned short
  18.             printf("%llu,%hu,%hu\n",sizeof(unsigned short),0,USHRT_MAX);
  19.             break;
  20.         case 5: // int
  21.             printf("%llu,%d,%d\n",sizeof(int),INT_MIN,INT_MAX);
  22.             break;
  23.         case 6: // unsigned int
  24.             printf("%llu,%u,%u\n",sizeof(unsigned int),0,UINT_MAX);
  25.             break;
  26.         case 7: //long
  27.             printf("%llu,%ld,%ld\n",sizeof(int),LONG_MIN,LONG_MAX);
  28.             break;
  29.         case 8: // unsigned long
  30.             printf("%llu,%u,%lu\n",sizeof(unsigned long),0,ULONG_MAX);
  31.             break;
  32.         case 9: // long long
  33.             printf("%llu,%lld,%lld\n",sizeof(int),LLONG_MIN,LLONG_MAX);
  34.             break;
  35.         case 10: // unsigned long long
  36.             printf("%llu,%u,%llu\n",sizeof(unsigned long long),0,ULLONG_MAX);
  37.             break;
  38.     }
  39.     return 0;
  40. }

平均值

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     int a = 0, b = 0;
  5.     scanf("%d %d",&a,&b);
  6.     int c = ((b-a)>>1)+a;
  7.     printf("%d",c);
  8.     return 0;
  9. }

进制转换

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     unsigned int a = 0;
  5.     scanf("%d",&a);
  6.     printf("%X,%o",a,a);
  7. }

浮点数输出

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     double a = 0.0f;
  5.     scanf("%lf",&a);
  6.     printf("%.6lf,%.2lf,%.8lf",a,a,a);
  7.     return 0;
  8. }

动态宽度输出

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     int n = 0, m = 0, k = 0;
  5.     scanf("%d %d",&n,&m);
  6.     int tmp = n;
  7.     while (tmp) {
  8.         tmp /= 10;
  9.         ++k;
  10.     }
  11.     for (int i = 0; i < m-k; ++i) {
  12.         printf("%d",0);
  13.     }
  14.     printf("%d",n);
  15.     return 0;
  16. }

计算地球上两点之间的距离

  1. #include <stdio.h>
  2. #include <math.h>
  3. #define RADIUS 6371.000000
  4. #define PI 3.1415926
  5.  
  6. int main(){
  7.     double phi1, phi2, lambda1, lambda2, distance;
  8.     scanf("%lf %lf",&phi1,&lambda1);
  9.     scanf("%lf %lf",&phi2,&lambda2);
  10.     phi1 = phi1*PI/180;
  11.     phi2 = phi2*PI/180;
  12.     lambda1 = lambda1*PI/180;
  13.     lambda2 = lambda2*PI/180;
  14.     double havRatio = (1-cos(phi2-phi1))/2+cos(phi1)*cos(phi2)*(1-cos(lambda2-lambda1))/2;
  15.     distance = asin(sqrt(havRatio))*2*RADIUS;
  16.     printf("%.4lfkm",distance);
  17.     return 0;
  18. }

风寒指数

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. int main(){
  5.     double v,T;
  6.     scanf("%lf %lf",&v,&T);
  7.     int chill = 13.12f+0.6215f*T-11.37f*pow(v,0.16f)+0.3965f*T*pow(v,0.16f)+0.5f;
  8.     printf("%d",chill);
  9.     return 0;
  10. }

颜色模型转换

  1. #include <stdio.h>
  2. #include <stdio.h>
  3. #include <math.h>
  4.  
  5. double findmax(double a, double b, double c) {
  6.     return a >= b ? (a >= c ? a : c) : (b >= c ? b : c);
  7. }
  8.  
  9. double findmin(double a, double b, double c) {
  10.     return a <= b ? (a <= c ? a : c) : (b <= c ? b : c);
  11. }
  12.  
  13. int main()
  14. {
  15.     int R, G, B;
  16.     double r, g, b;
  17.     double max, min;
  18.     double H, S, V;
  19.  
  20.     scanf("%d %d %d",&R,&G,&B);
  21.  
  22.     r = (double)((R > 0 ? R : 0) / 255.0f);
  23.     g = (double)((G > 0 ? G : 0) / 255.0f);
  24.     b = (double)((B > 0 ? B : 0) / 255.0f);
  25.     max = findmax(r, g, b);
  26.     min = findmin(r, g, b);
  27.  
  28.     V = max;
  29.  
  30.     if (max < 1e-9) {
  31.         S = 0.0f;
  32.     }
  33.     else {
  34.         S = (max - min) / max;
  35.     }
  36.  
  37.     if (max - min < 1e-9) {
  38.         H = 0.0f;
  39.     }
  40.     else {
  41.         if (max == r) {
  42.             H = 60.0f * (g - b) / (max - min);
  43.         }
  44.         else if (max == g) {
  45.             H = 60.0f * (2.0f + (b - r) / (max - min));
  46.         }
  47.         else if (max == b) {
  48.             H = 60.0f * (4.0f + (r - g) / (max - min));
  49.         }
  50.         else {
  51.             return -1;
  52.         }
  53.  
  54.         if (H < 1e-9) {
  55.             H = H + 360.0f;
  56.         }
  57.     }
  58.     printf("%.4lf,%.4lf%%,%.4lf%%", H, S * 100, V * 100);
  59.     return 0;
  60. }

11-20

乘数模

  1. #include<stdio.h>
  2.  
  3. int main(){
  4.     long long a,b,m,r;
  5.     scanf("%lld %lld %lld",&a,&b,&m);
  6.     r = ((a%m)*(b%m))%m;
  7.     printf("%lld",r);
  8.     return 0;
  9. }

方阵

  1. #include<stdio.h>
  2.  
  3. int main(){
  4.     int n,m;
  5.     scanf("%d",&n);
  6.     for (int i = 0; i < n; ++i) {
  7.         for (int j = 0; j < n; ++j) {
  8.             m = (i-j) > 0 ? (i-j) : (j-i);
  9.             printf("%d  ",m);
  10.         }
  11.         printf("\n");
  12.     }
  13.  
  14.     return 0;
  15. }

分数的加、减、乘、除法

  1. #include<stdio.h>
  2.  
  3. int gcd(int a,int b){
  4.     if (b == 0) return a;
  5.     else return gcd(b,a%b);
  6. }
  7.  
  8. int main(){
  9.     int aTop, aBot, bTop, bBot;
  10.     scanf("%d",&aTop);
  11.     getchar();
  12.     scanf("%d",&aBot);
  13.     getchar();
  14.     scanf("%d",&bTop);
  15.     getchar();
  16.     scanf("%d",&bBot);
  17.     getchar();
  18.  
  19.     int addTop, addBot, subTop, subBot, multTop, multBot, divTop, divBot;
  20.     int addGcd, subGcd, multGcd, divGcd;
  21.  
  22.     addTop = aTop*bBot+aBot*bTop;
  23.     addBot = aBot*bBot;
  24.     addGcd = gcd(addTop,addBot);
  25.     addTop /= addGcd;
  26.     addBot /= addGcd;
  27.  
  28.  
  29.     subTop = aTop*bBot-aBot*bTop;
  30.     subBot = aBot*bBot;
  31.     subGcd = gcd(subTop,subBot);
  32.     subTop /= subGcd;
  33.     subBot /= subGcd;
  34.  
  35.     multTop = aTop*bTop;
  36.     multBot = aBot*bBot;
  37.     multGcd = gcd(multTop,multBot);
  38.     multTop /= multGcd;
  39.     multBot /= multGcd;
  40.  
  41.     divTop = aTop*bBot;
  42.     divBot = aBot*bTop;
  43.     divGcd = gcd(divTop,divBot);
  44.     divTop /= divGcd;
  45.     divBot /= divGcd;
  46.  
  47.     printf("(%d/%d)+(%d/%d)=%d/%d\n",aTop,aBot,bTop,bBot,addTop,addBot);
  48.     printf("(%d/%d)-(%d/%d)=%d/%d\n",aTop,aBot,bTop,bBot,subTop,subBot);
  49.     printf("(%d/%d)*(%d/%d)=%d/%d\n",aTop,aBot,bTop,bBot,multTop,multBot);
  50.     printf("(%d/%d)/(%d/%d)=%d/%d\n",aTop,aBot,bTop,bBot,divTop,divBot);
  51.  
  52.     return 0;
  53. }

操作数

  1. #include<stdio.h>
  2.  
  3. int sumDig(int n){
  4.     int sum = 0;
  5.     while(n){
  6.         sum += n%10;
  7.         n /= 10;
  8.     }
  9.     return sum;
  10.  
  11. }
  12.  
  13. int main(){
  14.     int n,cnt=0;
  15.     scanf("%d",&n);
  16.  
  17.     while(n){
  18.         n -= sumDig(n);
  19.         ++cnt;
  20.     }
  21.  
  22.     printf("%d",cnt);
  23.  
  24.     return 0;
  25. }

组合数

  1. #include<stdio.h>
  2.  
  3. int main(){
  4.  
  5.     int cnt = 0, n, m;
  6.     scanf("%d",&n);
  7.  
  8.     for(int a = 0; a <= 9; ++a){
  9.         for(int b = 0; b <= 9; ++b){
  10.             for(int c = 0; c <= 9; ++c){
  11.                 for(int d = 0; d <= 9; ++d){
  12.                     m = a+b+c+d;
  13.                     if (m == n){
  14.                         ++cnt;
  15.                     }
  16.                 }
  17.  
  18.             }
  19.         }
  20.     }
  21.  
  22.     printf("%d",cnt);
  23.  
  24.     return 0;
  25. }

比率

  1. #include<stdio.h>
  2. #include<math.h>
  3.  
  4. int gcd(int a,int b){
  5.     if (b == 0) return a;
  6.     else return gcd(b,a%b);
  7. }
  8.  
  9. int cntDig(double x){
  10.     int cnt = 0;
  11.     while(x != floor(x)){
  12.         ++cnt;
  13.         x *= 10;
  14.     }
  15.     return cnt;
  16. }
  17.  
  18. int main(){
  19.  
  20.     double x;
  21.     scanf("%lf",&x);
  22.  
  23.     int cnt = cntDig(x), n, m, g;
  24.     m = pow(10,cnt);
  25.     n = (int)(x*m);
  26.     g = gcd(m,n);
  27.     m /= g;
  28.     n /= g;
  29.  
  30.     printf("%d/%d",n,m);
  31.     return 0;
  32. }

级数和

  1. #include<stdio.h>
  2. #include<math.h>
  3.  
  4. double decimal (int n){
  5.     double m = (double)n+1;
  6.     while(floor(m)){
  7.         m /= 10;
  8.     }
  9.     return m+n;
  10. }
  11.  
  12. int main(){
  13.     int n;
  14.     scanf("%d",&n);
  15.     double sum = 0.0f, term;
  16.     for (int i = 1; i < n; ++i) {
  17.         term = decimal(i);
  18.         sum += term;
  19.         printf("%g+",term);
  20.     }
  21.     term = decimal(n);
  22.     sum += term;
  23.     printf("%g=%g",term,sum);
  24.  
  25.     return 0;
  26. }

对称数

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.  
  5.     int n;
  6.     scanf("%d",&n);
  7.  
  8.     int a = (n/100)%10;
  9.     int b = (n/10)%10;
  10.     int c = n%10;
  11.  
  12.     if ((b == 0 || b == 1 || b == 8) && ((a == 6 && c == 9) || (a == 9 && c == 6) || ((a == 0 || a == 1 || a == 8) && (c == 0 || c == 1 || c == 8)))){
  13.         printf("Yes\n");
  14.     } else {
  15.         printf("No\n");
  16.     }
  17.  
  18.     return 0;
  19. }

幂数模

  1. #include<stdio.h>
  2.  
  3. typedef unsigned long long uint64;
  4.  
  5. int fastPowerMod (uint64 t, uint64 e, uint64 m){
  6.     uint64 r = 1;
  7.     while (e){
  8.         if (e&1){
  9.             r = (1LL*r*t)%m;
  10.         }
  11.         t = (1LL*t*t)%m;
  12.         e >>= 1;
  13.     }
  14.     return r;
  15. }
  16.  
  17. int main(){
  18.     uint64 a,b,m,r;
  19.     scanf("%llu %llu %llu",&a,&b,&m);
  20.     r = fastPowerMod(a,b,m);
  21.     printf("%llu",r);
  22.     return 0;
  23. }

倍数和

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     unsigned int t = 0;
  5.     unsigned int arrn[100000];
  6.     scanf("%u", &t);
  7.     for(unsigned int i = 0; i < t; i++){
  8.         scanf("%u", &arrn[i]);
  9.     }
  10.  
  11.     for(unsigned int j = 0; j < t; j++){
  12.         unsigned int res = 0;
  13.         for(unsigned int x = 1; x < arrn[j]; x++){
  14.             if(x % 3 == 0 || x % 5 == 0){
  15.                 res += x;
  16.             }
  17.         }
  18.         printf("%u\n", res);
  19.     }
  20.  
  21.     return 0;
  22. }

21-30

余数和

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     unsigned int sum = 0, n, k;
  5.     scanf("%u %u",&n,&k);
  6.     for (unsigned int i = 1; i <= n; ++i) {
  7.         sum += k%i;
  8.     }
  9.     printf("%u",sum);
  10.     return 0;
  11. }

最大数字

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. bool isNonDecr(unsigned int n){
  5.     unsigned int left;
  6.     while(n){
  7.         left = (n/10)%10;
  8.         if (left > (n%10)){
  9.             return false;
  10.         }
  11.         n /= 10;
  12.     }
  13.     return true;
  14. }
  15.  
  16. int main() {
  17.     unsigned int n, res;
  18.     scanf("%u",&n);
  19.     while(!isNonDecr(n)){
  20.         --n;
  21.     }
  22.     printf("%u",n);
  23.     return 0;
  24. }

倒水（原 AC 后 TE 现又 AC）

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. #define INF 0x3f3f3f3f
  5.  
  6. const int maxSize = 10000;
  7. int distTable[10000][10000];
  8. int queue[2][10000] = {0};
  9.  
  10. int bfs(int n, int m, int d){
  11.     int head = 0, tail = 1;
  12.     memset(distTable, 0x3f, sizeof(distTable));
  13.     distTable[0][0] = 0;
  14.  
  15.     while(tail != head){
  16.         int a = queue[0][head], b = queue[1][head];
  17.         if (a == d || b == d) {
  18.             return distTable[a][b];
  19.         }
  20.  
  21.         ++head;
  22.         if(head == maxSize-1) {
  23.             head = 0;
  24.         }
  25.  
  26.         // full
  27.         if(distTable[a][m] == INF) {
  28.             distTable[a][m] = distTable[a][b]+1;
  29.             ++tail;
  30.             queue[0][tail] = a;
  31.             queue[1][tail] = m;
  32.         }
  33.         if(distTable[n][b] == INF) {
  34.             distTable[n][b] = distTable[a][b]+1;
  35.             ++tail;
  36.             queue[0][tail] = n;
  37.             queue[1][tail] = b;
  38.         }
  39.  
  40.         // empty
  41.         if(distTable[a][0] == INF) {
  42.             distTable[a][0] = distTable[a][b]+1;
  43.             ++tail;
  44.             queue[0][tail] = a;
  45.             queue[1][tail] = 0;
  46.         }
  47.         if(distTable[0][b] == INF) {
  48.             distTable[0][b] = distTable[a][b]+1;
  49.             ++tail;
  50.             queue[0][tail] = 0;
  51.             queue[1][tail] = b;
  52.         }
  53.  
  54.         // pour
  55.         if(a+b <= n && distTable[a+b][0] == INF){
  56.             distTable[a+b][0] = distTable[a][b]+1;
  57.             ++tail;
  58.             queue[0][tail] = a+b;
  59.             queue[1][tail] = 0;
  60.         }
  61.         if(a+b > n && distTable[n][a+b-n] == INF){
  62.             distTable[n][a+b-n] = distTable[a][b]+1;
  63.             ++tail;
  64.             queue[0][tail] = n;
  65.             queue[1][tail] = a+b-n;
  66.         }
  67.         if(a+b <= m && distTable[0][a+b] == INF){
  68.             distTable[0][a+b] = distTable[a][b]+1;
  69.             ++tail;
  70.             queue[0][tail] = 0;
  71.             queue[1][tail] = a+b;
  72.         }
  73.         if(a+b > m && distTable[a+b-m][m] == INF){
  74.             distTable[a+b-m][m] = distTable[a][b]+1;
  75.             ++tail;
  76.             queue[0][tail] = a+b-m;
  77.             queue[1][tail] = m;
  78.         }
  79.  
  80.         if (tail == maxSize) {
  81.             tail = 0;
  82.         }
  83.     }
  84.  
  85.     return -1;
  86. }
  87.  
  88. int main(){
  89.     int n,m,d;
  90.     scanf("%d %d %d",&n,&m,&d);
  91.     printf("%d", bfs(n,m,d));
  92.     return 0;
  93. }

好数字

  1. #include <stdio.h>
  2.  
  3. typedef unsigned long long uint64;
  4. const uint64 mod = 1e9+7;
  5.  
  6. uint64 power(uint64 a, uint64 e){
  7.     uint64 r = 1;
  8.     while (e){
  9.         if (e&1){
  10.             r = (r*a)%mod;
  11.         }
  12.         a = (a*a)%mod;
  13.         e >>= 1;
  14.     }
  15.     return r;
  16. }
  17.  
  18. int main(){
  19.     uint64 n, num;
  20.     scanf("%llu",&n);
  21.     num = power(4,n/2)*power(5,n-n/2)%mod;
  22.     printf("%llu",num);
  23.     return 0;
  24. }

毕达哥拉斯三元组

  1. #include <stdio.h>
  2.  
  3. unsigned long long pythagoras(unsigned int sum){
  4.     // a as short catheti, b as long catheti, c as hypotenuse
  5.     unsigned a,b,c;
  6.     for (a = 1; a <= sum/4; ++a){
  7.         for (b = a+1; b <= sum/2; ++b){
  8.             c = sum-a-b;
  9.             if ((a+b > c) & (c-a < b) & (a*a+b*b == c*c)){
  10.                 return a*b*c;
  11.             }
  12.         }
  13.     }
  14. }
  15.  
  16. int main(){
  17.     unsigned int n;
  18.     scanf("%u",&n);
  19.     printf("%d", pythagoras(n));
  20.     return 0;
  21. }

竖式乘法

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. typedef unsigned int uint;
  5.  
  6. uint digit(uint x){
  7.     uint cnt = 0;
  8.     if (x == 0){
  9.         return 1;
  10.     }
  11.     while(x){
  12.         x /= 10;
  13.         ++cnt;
  14.     }
  15.     return cnt;
  16. }
  17.  
  18. uint getDigital(uint x, uint n){
  19.     x /= (uint)pow(10,n-1);
  20.     return x%10;
  21. }
  22.  
  23. int main(){
  24.     uint up, down;
  25.     scanf("%u %u",&up,&down);
  26.     uint ans = up*down, len = digit(ans)+1, downLen = digit(down);
  27.  
  28.     for (uint i = 0; i < len-digit(up); ++i) {
  29.         printf(" ");
  30.     }
  31.     printf("%u\n",up);
  32.     printf("x");
  33.     for (uint i = 1; i <= len-downLen-1; ++i) {
  34.         printf(" ");
  35.     }
  36.     printf("%u\n",down);
  37.  
  38.     for (uint i = 1; i <= len; ++i) {
  39.         printf("-");
  40.     }
  41.     printf("\n");
  42.  
  43.     uint tmp;
  44.     for (uint i = 1; i <= downLen; ++i) {
  45.         tmp = getDigital(down,i)*up;
  46.         if (i == downLen){
  47.             printf("+");
  48.         } else {
  49.             for (uint j = 1; j <= len-digit(tmp)-i+1; ++j) {
  50.                 printf(" ");
  51.             }
  52.         }
  53.         printf("%u\n",tmp);
  54.     }
  55.  
  56.     for (uint i = 1; i <= len; ++i) {
  57.         printf("-");
  58.     }
  59.     printf("\n");
  60.  
  61.     printf(" %u",ans);
  62.  
  63.     return 0;
  64. }

查找数列

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     int cnt = 1, sum = 0, n, res;
  5.     scanf("%d",&n);
  6.     while(n-sum > 0){
  7.         sum += cnt;
  8.         ++cnt;
  9.     }
  10.     sum -= cnt;
  11.     res = n-sum == 0 ? cnt : (n-sum-1);
  12.     printf("%d",res);
  13.     return 0;
  14. }

俄罗斯农夫乘法

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     int multi = 0, a, b;
  5.     scanf("%d %d",&a,&b);
  6.     while(a){
  7.         printf("%d %d\n",a,b);
  8.         if (a&1) {
  9.             multi += b;
  10.         }
  11.         a >>= 1;
  12.         b <<= 1;
  13.     }
  14.     printf("%d",multi);
  15.     return 0;
  16. }

阶乘倍数

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. typedef unsigned long long int64;
  5.  
  6. int64 primeFactNum = 0;
  7. int64 prime[20] = {0}, num[20] = {0};
  8.  
  9. bool isMulti(int64 n){
  10.     int64 primeNum, tmp;
  11.     for (int64 i = 1; i <= primeFactNum; ++i) {
  12.         primeNum = 0;
  13.         tmp = n;
  14.         while (tmp) {
  15.             primeNum += tmp/prime[i];
  16.             tmp /= prime[i];
  17.         }
  18.         if(primeNum < num[i]) {
  19.             return false;
  20.         }
  21.     }
  22.     return true;
  23. }
  24.  
  25. void solveFact(int64 k){
  26.     for (int64 i = 2; i*i <= k; ++i) {
  27.         if(k%i == 0){
  28.             ++primeFactNum;
  29.             prime[primeFactNum] = i;
  30.             while (k%i == 0){
  31.                 ++num[primeFactNum];
  32.                 k /= i;
  33.             }
  34.         }
  35.     }
  36.     if (k > 1){
  37.         ++primeFactNum;
  38.         prime[primeFactNum] = k;
  39.         ++num[primeFactNum];
  40.     }
  41. }
  42.  
  43. int main(){
  44.     int64 left = 1, right = 1e19, mid, n, k;
  45.     scanf("%lld",&k);
  46.     solveFact(k);
  47.  
  48.     while(left <= right){
  49.         mid = ((right-left)>>1)+left;
  50.         if (isMulti(mid)){
  51.             right = mid-1;
  52.             n = mid;
  53.         } else {
  54.             left = mid+1;
  55.         }
  56.     }
  57.     printf("%lld",n);
  58.     return 0;
  59. }

方案数

  1. #include <stdio.h>
  2.  
  3. int main(){
  4.     int cnt = 0, n;
  5.     scanf("%d",&n);
  6.     for (int i = 1; i*(i+1) <= 2*n; ++i) {
  7.         if ((n-i*(i-1)/2)%i == 0){
  8.             ++cnt;
  9.         }
  10.     }
  11.     printf("%d",cnt);
  12.     return 0;
  13. }

31-40

哈沙德数

  1. #include <stdio.h>
  2.  
  3. int HarshadNumber(int n){
  4.     int t = n, s = 0;
  5.     while (t) {
  6.         s += t%10;
  7.         t /= 10;
  8.     }
  9.     if ((s == 0) || (n%s != 0)) return 0;
  10.     if (s == 1) return 1;
  11.     return n/s;
  12. }
  13.  
  14. int main(){
  15.     int cnt = 0, n;
  16.     scanf("%d",&n);
  17.     if (n == 1) cnt = 1;
  18.     while ((n != 0) && (n != 1)) {
  19.         n = HarshadNumber(n);
  20.         if (n) ++cnt;
  21.     }
  22.     printf("%d",cnt);
  23.     return 0;
  24. }

素数

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3. #include <string.h>
  4.  
  5. typedef unsigned long long uint64;
  6.  
  7. uint64 primeNum(uint64 a, uint64 b){
  8.     bool isPrime[b+1];
  9.     memset(isPrime,1,b+1);
  10.     uint64 cnt = 0;
  11.     for (uint64 i = 2; i <= b; ++i){
  12.         if (isPrime[i]){
  13.             for (uint64 j = 2; j*i <= b; ++j){
  14.                 isPrime[j*i] = false;
  15.             }
  16.         }
  17.     }
  18.     for (uint64 i = a; i <= b; ++i){
  19.         cnt += isPrime[i];
  20.     }
  21.     return cnt;
  22. }
  23.  
  24. int main(){
  25.     uint64 a,b,num;
  26.     scanf("%llu %llu",&a,&b);
  27.     num = primeNum(a,b);
  28.     printf("%llu",num);
  29.     return 0;
  30. }

基思数

  1. #include <stdio.h>
  2.  
  3. int arr[8] = {0};
  4.  
  5. int init(int n){
  6.     int cnt = 0;
  7.     while (n) {
  8.         arr[cnt++] = n%10;
  9.         n /= 10;
  10.     }
  11.     return cnt;
  12. }
  13.  
  14. void isKeith(int n, int len){
  15.     int i = len - 1;
  16.     while (arr[i] < n){
  17.         int sum = 0;
  18.         for (int j = 0; j < len; ++j) {
  19.             sum += arr[(i-j+len)%len];
  20.         }
  21.         arr[i] = sum;
  22.         i = (i-1+len)%len;
  23.     }
  24.     if (arr[i] == n) printf("Yes");
  25.     else printf("No");
  26. }
  27.  
  28. int main() {
  29.     int n;
  30.     scanf("%d",&n);
  31.     isKeith(n,init(n));
  32.     return 0;
  33. }

二进制表示

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. void binary(int a){
  5.     bool flag = false;
  6.     for (int i = 15; i >= 0 ; --i) {
  7.         if ((a>>i)&1) {
  8.             if (flag) printf("+");
  9.             if (i >= 2){
  10.                 printf("2(");
  11.                 binary(i);
  12.                 printf(")");
  13.             }
  14.             if (i == 1) printf("2");
  15.             if (i == 0) printf("2(0)");
  16.             flag = true;
  17.         }
  18.     }
  19. }
  20.  
  21. int main() {
  22.     int a;
  23.     scanf("%d",&a);
  24.     binary(a);
  25.     return 0;
  26. }

光线追踪

  1. #include <stdio.h>
  2.  
  3. unsigned int gcd(unsigned int a, unsigned int b){
  4.     if (b == 0) return a;
  5.     return gcd(b,a%b);
  6. }
  7.  
  8. int main(){
  9.     unsigned int n,x,l;
  10.     scanf("%u %u",&n,&x);
  11.     l = 3*(n-gcd(n,x));
  12.     printf("%u",l);
  13.     return 0;
  14. }

冰雹数列

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     int n;
  5.     scanf("%d",&n);
  6.     while (n != 1) {
  7.         printf("%d ",n);
  8.         if (n&1) n = 3 * n + 1;
  9.         else n /= 2;
  10.     }
  11.     printf("1");
  12.     return 0;
  13. }

佩尔数

  1. #include <stdio.h>
  2.  
  3. int PA(int n){
  4.     if (n == 0) return 0;
  5.     else if (n == 1) return 1;
  6.     return 2*PA(n-1)+PA(n-2);
  7. }
  8.  
  9. int PB(int n){
  10.     int p0 = 0, p1 = 1, pn;
  11.     for (int i = 0; i <= n; ++i) {
  12.         if (i == 0) pn = p0;
  13.         else if (i == 1) pn = p1;
  14.         else {
  15.             pn = 2 * p1 + p0;
  16.             p0 = p1;
  17.             p1 = pn;
  18.         }
  19.     }
  20.     return pn;
  21. }
  22.  
  23. int main() {
  24.     int n, p;
  25.     scanf("%d",&n);
  26.     if (n&1) p = PA(n);
  27.     else p = PB(n);
  28.     printf("%d",p);
  29.     return 0;
  30. }

可变参数累加

  1. #include <stdio.h>
  2. #include <stdarg.h>
  3.  
  4. int sum(int start,...){
  5.     va_list vaList;
  6.     int sum = 0;
  7.     int curr = start;
  8.     va_start(vaList,start);
  9.     while (start){
  10.         sum += start;
  11.         start = va_arg(vaList,int);
  12.     }
  13.     va_end(vaList);
  14.     return sum;
  15. }
  16.  
  17. int main() {
  18.     int a,b,c,d,e,f;
  19.     scanf("%d %d %d %d %d %d",&a,&b,&c,&d,&e,&f);
  20.     int sumMinus = sum(a,b,0) - sum(c,d,e,f,0);
  21.     printf("%d",sumMinus);
  22.     return 0;
  23. }

运动会

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. int phiEuler(int n){
  5.     int phi[n+1],prime[n+1];
  6.     bool isSieved[n+1];
  7.     int sum = 0,cnt = 1, comp;
  8.     prime[0] = 1;
  9.     phi[1] = 1;
  10.     for (int i = 2; i < n; ++i){
  11.         if (!isSieved[i]){
  12.             prime[cnt++] = i;
  13.             phi[i] = i-1;
  14.         }
  15.         for (int j = 1; i*prime[j] <= n; ++j){
  16.             comp = i*prime[j];
  17.             isSieved[comp] = true;
  18.             if (i%prime[j] == 0){
  19.                 phi[comp] = prime[j]*phi[i];
  20.                 break;
  21.             } else{
  22.                 phi[comp] = (prime[j]-1)*phi[i];
  23.             }
  24.         }
  25.     }
  26.     for (int i = 1; i <= n-1; ++i) {
  27.         sum += phi[i];
  28.     }
  29.     return sum;
  30. }
  31.  
  32. int main() {
  33.     int n, num;
  34.     scanf("%d",&n);
  35.     num = n == 1 ? 0 : (2*phiEuler(n)+1);
  36.     printf("%d",num);
  37.     return 0;
  38. }

可变参数平均

  1. #include <stdio.h>
  2. #include <stdarg.h>
  3.  
  4. double avg(int num,...){
  5.     va_list vaList;
  6.     double sum = 0.0f;
  7.     va_start(vaList,num);
  8.     for (int i = 0; i < num; ++i) {
  9.         sum += va_arg(vaList,int);
  10.     }
  11.     va_end(vaList);
  12.     return sum/num;
  13. }
  14.  
  15. int main() {
  16.     int a,b,c,d,e;
  17.     scanf("%d %d %d %d %d",&a,&b,&c,&d,&e);
  18.     double avgMinus = avg(2,a,b) - avg(3,c,d,e);
  19.     printf("%.4f",avgMinus);
  20.     return 0;
  21. }

41-50

航空旅行

  1. #include<stdio.h>
  2. #include<stdbool.h>
  3.  
  4. void pass(int a, int b, int c, int d, int e){
  5.     bool flag = false;
  6.     if (a <= e && (b + c) <= d) flag = true;
  7.     if (b <= e && (a + c) <= d) flag = true;
  8.     if (c <= e && (a + b) <= d) flag = true;
  9.     if (flag) printf("YES\n");
  10.     else printf("NO\n");
  11. }
  12.  
  13. int main(){
  14.     int n, a, b, c, d, e;
  15.     scanf("%d",&n);
  16.     while (n--){
  17.         scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);
  18.         pass(a, b, c, d, e);
  19.     }
  20.     return 0;
  21. }

蒙特卡罗法求积分

  1. #include <stdio.h>
  2. #include <math.h>
  3. #include <stdlib.h>
  4.  
  5. double func1(double x) {
  6.     return pow(x, 4) * exp(-x);
  7. }
  8.  
  9. double func2(double x) {
  10.     return x * x + 1;
  11. }
  12.  
  13. double func3(double x) {
  14.     return cos(x);
  15. }
  16.  
  17. double func4(double x) {
  18.     return sqrt(x) * (x - 2);
  19. }
  20.  
  21. double func5(double x) {
  22.     return 2 * sin(x) - 5 * cos(x);
  23. }
  24.  
  25. double func(int m, double x) {
  26.     switch (m) {
  27.         case 1: return func1(x);
  28.         case 2: return func2(x);
  29.         case 3: return func3(x);
  30.         case 4: return func4(x);
  31.         case 5: return func5(x);
  32.         default: return 0;
  33.     }
  34. }
  35.  
  36. double mtk(int m, double a, double b, int n) {
  37.     srand(RAND_MAX);
  38.     double w = b - a, sum = 0;
  39.     for (int i = 1; i < n; ++i) {
  40.         double x = ((double)rand() / RAND_MAX) * w + a;
  41.         sum += func(m, x);
  42.     }
  43.     sum *= w / n;
  44.     return sum;
  45. }
  46.  
  47. int main() {
  48.     int m, n;
  49.     double a, b;
  50.     scanf("%d %lf %lf %d", &m, &a, &b, &n);
  51.     printf("%.6lf", mtk(m, a, b, n));
  52.     return 0;
  53. }

飞机起飞速度（WA）

稀疏矩阵

  1. #include <stdio.h>
  2.  
  3. int main () {
  4.     int raw, col, n, num = 0;
  5.     scanf("%d %d", &raw, &col);
  6.     for (int i = 0; i < raw; ++i) {
  7.         for (int j = 0; j < col; ++j) {
  8.             scanf("%d", &n);
  9.             if (n) ++num;
  10.         }
  11.     }
  12.     double ratio = (double)num / (raw * col);
  13.     if (num == raw || num == col || (ratio - 0.05) <= 1e-9)
  14.         printf("Yes\n");
  15.     else printf("No\n");
  16.     return 0;
  17. }

回文数之和

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. int dec[10] = {0}, kSys[32] = {0};
  5.  
  6. bool isPalindrome(int arr[], int cnt){
  7.     int head = 0, tail = cnt - 1;
  8.     while (head < tail) {
  9.         if (arr[head] != arr[tail]) return false;
  10.         ++head, --tail;
  11.     }
  12.     return true;
  13. }
  14.  
  15. bool isBiPalindrome(int n, int k){
  16.     int tmp = n, cnt = 0;
  17.     while (tmp) {
  18.         dec[cnt++] = tmp % 10;
  19.         tmp /= 10;
  20.     }
  21.     if (!isPalindrome(dec, cnt)) return false;
  22.  
  23.     tmp = n, cnt = 0;
  24.     while (tmp) {
  25.         kSys[cnt++] = tmp % k;
  26.         tmp /= k;
  27.     }
  28.     if (!isPalindrome(kSys, cnt)) return false;
  29.     return true;
  30. }
  31.  
  32. int main() {
  33.     int n, k, sum = 0;
  34.     scanf("%d %d", &n, &k);
  35.     for (int i = 1; i <= n; ++i) {
  36.         if (isBiPalindrome(i, k)) sum += i;
  37.     }
  38.     printf("%d", sum);
  39.     return 0;
  40. }

完美矩阵

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. #define MAXSIZE 301
  5.  
  6. int arr[MAXSIZE][MAXSIZE] = {0};
  7. int preSum[MAXSIZE][MAXSIZE] = {0};
  8.  
  9. void prefix(int n, int m){
  10.     for (int i = 1; i <= n; ++i) {
  11.         for (int j = 1; j <= m; ++j) {
  12.             preSum[i][j] = preSum[i - 1][j] + preSum[i][j - 1]
  13.                            - preSum[i - 1][j - 1] + arr[i][j];
  14.         }
  15.     }
  16. }
  17.  
  18. int getSum(int x1, int x2, int y1, int y2) {
  19.     return preSum[x2][y2] - preSum[x1 - 1][y2] - preSum[x2][y1 - 1]
  20.            + preSum[x1 - 1][y1 - 1];
  21. }
  22.  
  23. bool isPerfect(int x1, int x2, int y1, int y2) {
  24.     int outer = getSum(x1, x2, y1, y2), inner;
  25.     int len = 2 * (x2 - x1 + y2 - y1);
  26.     if ((x2 - x1) == 1 || (y2 - y1) == 1) inner = 0;
  27.     else inner = getSum(x1 + 1, x2 - 1, y1 + 1, y2 - 1);
  28.  
  29.     if (inner != 1 && inner != 0 && inner != -1) return false;
  30.     if ((outer - inner) != len) return false;
  31.     return true;
  32. }
  33.  
  34. int perfectNum(int n, int m) {
  35.     int cnt = 0;
  36.     for (int i = 1; i <= n; ++i) {
  37.         for (int j = 1; j <= m; ++j) {
  38.             for (int k = 1; k + i <= n && k + j <= m; ++k) {
  39.                 if (arr[i][k + j] == 0 || arr[k + i][j] == 0) break;
  40.                 if (isPerfect(i, i + k, j, j + k)) {
  41.                     ++cnt;
  42.                 }
  43.             }
  44.         }
  45.     }
  46.     return cnt;
  47. }
  48.  
  49. int main () {
  50.     int n, m;
  51.     scanf("%d %d", &n, &m);
  52.     for (int i = 1; i <= n; ++i) {
  53.         for (int j = 1; j <= m; ++j) {
  54.             scanf("%d", &arr[i][j]);
  55.             if (arr[i][j] == 0) arr[i][j] = -1;
  56.         }
  57.     }
  58.     prefix(n ,m);
  59.     printf("%d", perfectNum(n, m));
  60.     return 0;
  61. }

波士顿房价预测

  1. #include <stdio.h>
  2.  
  3. double nAvg(double arr[], int n) {
  4.     double sum = 0;
  5.     for (int i = 0; i < n; ++i) {
  6.         sum += arr[i];
  7.     }
  8.     return sum / n;
  9. }
  10.  
  11. int main() {
  12.     int n;
  13.     scanf("%d", &n);
  14.     double x[n], y[n];
  15.     for (int i = 0; i < n; ++i) {
  16.         scanf("%lf %lf", &x[i], &y[i]);
  17.     }
  18.  
  19.     double xBar = nAvg(x, n), yBar = nAvg(y, n);
  20.     double sumUp = 0, sumDown = 0;
  21.     for (int i = 0; i < n; ++i) {
  22.         sumUp += (x[i] - xBar) * (y[i] - yBar);
  23.     }
  24.     for (int i = 0; i < n; ++i) {
  25.         sumDown += (x[i] - xBar) * (x[i] - xBar);
  26.     }
  27.     double b = sumUp / sumDown;
  28.     double a = yBar - b * xBar;
  29.  
  30.     printf("Y=%.4lf+%.4lf*X",a,b);
  31.     return 0;
  32. }

行列式值

  1. #include <stdio.h>
  2.  
  3. #define MAX_SIZE 10
  4.  
  5. void swapRows(double matrix[MAX_SIZE][MAX_SIZE], int row1, int row2, int n) {
  6.     for (int i = 0; i < n; i++) {
  7.         double temp = matrix[row1][i];
  8.         matrix[row1][i] = matrix[row2][i];
  9.         matrix[row2][i] = temp;
  10.     }
  11. }
  12.  
  13. double calculateDeterminant(double matrix[MAX_SIZE][MAX_SIZE], int n) {
  14.     int i, j, k;
  15.     double determinant = 1.0;
  16.  
  17.     for (i = 0; i < n; i++) {
  18.         if (matrix[i][i] == 0.0) {
  19.             for (j = i + 1; j < n; j++) {
  20.                 if (matrix[j][i] != 0.0) {
  21.                     swapRows(matrix, i, j, n);
  22.                     determinant *= -1.0;
  23.                     break;
  24.                 }
  25.             }
  26.         }
  27.  
  28.         if (matrix[i][i] == 0.0) {
  29.             return 0.0;
  30.         }
  31.  
  32.         double pivot = matrix[i][i];
  33.         determinant *= pivot;
  34.         for (j = i; j < n; j++) {
  35.             matrix[i][j] /= pivot;
  36.         }
  37.  
  38.         for (j = i + 1; j < n; j++) {
  39.             double factor = matrix[j][i];
  40.             for (k = i; k < n; k++) {
  41.                 matrix[j][k] -= factor * matrix[i][k];
  42.             }
  43.         }
  44.     }
  45.  
  46.     return determinant;
  47. }
  48.  
  49. int main() {
  50.     int n;
  51.     scanf("%d", &n);
  52.  
  53.     double matrix[MAX_SIZE][MAX_SIZE];
  54.  
  55.     for (int i = 0; i < n; i++) {
  56.         for (int j = 0; j < n; j++) {
  57.             scanf("%lf", &matrix[i][j]);
  58.         }
  59.     }
  60.  
  61.     double determinant = calculateDeterminant(matrix, n);
  62.     printf("%.0lf\n", determinant);
  63.  
  64.     return 0;
  65. }
  1. #include <stdio.h>
  2. #include <stdlib.h>
  3.  
  4. int** init(int n) {
  5.     int** matrix = (int**)malloc(sizeof(int*) * (n + 1));
  6.     for(int i = 0; i <= n; i++) {
  7.         matrix[i] = (int*)malloc(sizeof(int) * (n + 1));
  8.     }
  9.     return matrix;
  10. }
  11.  
  12. int det(int **matrix, int n) {
  13.     if (n == 1) return matrix[1][1];
  14.     int sum = 0;
  15.     int **subMatrix = init(n - 1);
  16.     for (int i = 1; i <= n; ++i) {
  17.         for (int j = 1; j <= n - 1; ++j) {
  18.             for (int k = 1; k <= n - 1; ++k) {
  19.                 if (k < i) subMatrix[j][k] = matrix[j + 1][k];
  20.                 else subMatrix[j][k] = matrix[j + 1][k + 1];
  21.             }
  22.         }
  23.         int sgn = i % 2 == 0 ? -1 : 1;
  24.         sum += sgn * matrix[1][i] * det(subMatrix, n - 1);
  25.     }
  26.     return sum;
  27. }
  28.  
  29. int main() {
  30.     int n;
  31.     scanf("%d", &n);
  32.     int **matrix = init(n);
  33.     for (int i = 1; i <= n; ++i) {
  34.         for (int j = 1; j <= n; ++j) {
  35.             scanf("%d", &matrix[i][j]);
  36.         }
  37.     }
  38.     printf("%d", det(matrix, n));
  39.     return 0;
  40. }

货运优化

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     int l3s2[4] = {0, 5, 3, 1};
  5.     int n, x1, x2, x3, x4, x5, x6, s2, s1;
  6.     while (1) {
  7.         scanf("%d %d %d %d %d %d", &x1, &x2, &x3, &x4, &x5, &x6);
  8.         if ((x1 + x2 + x3 + x4 + x5 + x6) == 0) break;
  9.         n = (x3 + 3) / 4 + x4 + x5 + x6;
  10.         s2 = 5 * x4 + l3s2[x3 % 4];
  11.         if (x2 > s2) n += (x2 - s2 + 8) / 9;
  12.         s1 = 36 * n - 36 * x6 - 25 * x5 - 16 * x4 - 9 * x3 - 4 * x2;
  13.         if (x1 > s1) n += (x1 - s1 + 35) / 36;
  14.         printf("%d\n",n);
  15.     }
  16.     return 0;
  17. }

素数筛法

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. #define NUM (int)1e7+1
  5.  
  6. static bool isSieved[NUM];
  7. static int prime[NUM];
  8.  
  9. int main() {
  10.     int n, k = 0;
  11.     scanf("%d", &n);
  12.     isSieved[1] = true;
  13.     for (int i = 2; i <= n; ++i) {
  14.         if (!isSieved[i]) prime[++k] = i;
  15.         for (int j = 1; prime[j] * i <= n; ++j) {
  16.             isSieved[prime[j] * i] = true;
  17.             if (i % prime[j] == 0) break;
  18.         }
  19.     }
  20.     printf("%d", k);
  21. }

51-60

字符串替换（WA）

删除前后缀

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. void strRemovePrefix(char *str, char *prefix) {
  5.     int cnt = 0;
  6.     char *pStr = str, *pPrefix = prefix;
  7.     while (*pPrefix && *pStr && *pStr == *pPrefix) {
  8.         while (*pPrefix && *pStr && *pStr == *pPrefix)
  9.             ++pStr, ++pPrefix, ++cnt;
  10.         pPrefix = prefix;
  11.     }
  12.  
  13.     int len = strlen(prefix);
  14.     int mov = (cnt / len) * len;
  15.     if (mov) memmove(str - mov, str, strlen(str) + 1);
  16. }
  17.  
  18. void strRemoveSuffix(char *str, char *suffix) {
  19.     int len = strlen(suffix);
  20.     char *pStr = str + strlen(str) - len, *pSuffix = suffix;
  21.     while (*pSuffix && pStr >= str && *pStr == *pSuffix) {
  22.         int cnt = 0;
  23.         while (*pSuffix && pStr && *pStr == *pSuffix)
  24.             ++pStr, ++pSuffix, ++cnt;
  25.         if (cnt == len) {
  26.             pSuffix = suffix, pStr = pStr - 2 * len;
  27.             *(pStr + len) = '\0';
  28.         } else break;
  29.     }
  30. }
  31.  
  32. int main() {
  33.     char str1[1000] = "", fix[1000] = "", str2[1000] = "";
  34.     scanf("%[^\n] %[^\n]", str1, fix);
  35.     memcpy(str2, str1, strlen(str1) + 1);
  36.  
  37.     strRemovePrefix(str1, fix);
  38.     puts(str1);
  39.     strRemoveSuffix(str2, fix);
  40.     puts(str2);
  41.     return 0;
  42. }

大小写交换

  1. #include <stdio.h>
  2.  
  3. void strSwapCase(char str[]) {
  4.     for (int i = 0; str[i] != '\0'; ++i) {
  5.         if ('a' <= str[i] && str[i] <= 'z')
  6.             str[i] = (char) str[i] - 'a' + 'A';
  7.         else if ('A' <= str[i] && str[i] <= 'Z')
  8.             str[i] = (char) str[i] - 'A' + 'a';
  9.     }
  10. }
  11.  
  12. int main() {
  13.     char input[1000] = "";
  14.     scanf("%[^\n]",input);
  15.     strSwapCase(input);
  16.     puts(input);
  17.     return 0;
  18. }

前后缀移除

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. void strLeftStrip(char *str, char *chars) {
  5.     int mov = 0;
  6.     char *pStr = str;
  7.     while (*pStr) {
  8.         if (strchr(chars, *pStr)) ++mov;
  9.         else break;
  10.         ++pStr;
  11.     }
  12.     if (mov) memmove(str - mov, str, strlen(str) + 1);
  13. }
  14.  
  15. void strRightStrip(char *str, char *chars) {
  16.     int len = 0;
  17.     char *pStr = str + strlen(str) - 1;
  18.     while (pStr >= str) {
  19.         if (strchr(chars, *pStr)) ++len;
  20.         else break;
  21.         --pStr;
  22.     }
  23.     *(str + strlen(str) - len) = '\0';
  24. }
  25.  
  26. int main() {
  27.     char str1[1000] = "", chars[1000] = "", str2[1000] = "";
  28.     scanf("%[^\n] %[^\n]", str1, chars);
  29.     memcpy(str2, str1, strlen(str1) + 1);
  30.  
  31.     strLeftStrip(str1, chars);
  32.     puts(str1);
  33.     strRightStrip(str2, chars);
  34.     puts(str2);
  35.     strRightStrip(str1, chars);
  36.     puts(str1);
  37.     return 0;
  38. }

字符串后缀

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. void strEndsWith(char *str, char *suffix) {
  5.     int len = strlen(suffix);
  6.     char *pStr = str + strlen(str) - len, *pSuffix = suffix;
  7.     while (*pSuffix && *pStr) {
  8.         if (*pSuffix != *pStr) {
  9.             printf("No\n");
  10.             return;
  11.         }
  12.         else ++pSuffix, ++pStr;
  13.     }
  14.     printf("Yes\n");
  15. }
  16.  
  17. int main() {
  18.     char str[1000] = "", suffix[1000] = "";
  19.     scanf("%[^\n] %[^\n]", str, suffix);
  20.     strEndsWith(str, suffix);
  21.     return 0;
  22. }

Atol转换

  1. #include <stdio.h>
  2. #include <limits.h>
  3.  
  4. int atol(char *str) {
  5.     char *pStr = str;
  6.     int sgn = 1;
  7.     long long tmp = 0;
  8.     if (*pStr == '+') ++pStr;
  9.     else if (*pStr == '-') sgn = -1, ++pStr;
  10.  
  11.     while (*pStr) {
  12.         if (*pStr == ' ') ;
  13.         else if ('0' <= *pStr && *pStr <= '9') {
  14.             tmp = (*pStr - '0') + tmp * 10;
  15.             if ((tmp * sgn) >= INT_MAX) return INT_MAX;
  16.             else if ((tmp * sgn) <= INT_MIN) return INT_MIN;
  17.         }
  18.         else break;
  19.         ++pStr;
  20.     }
  21.     return tmp * sgn;
  22. }
  23.  
  24. int main() {
  25.     char str[1000] = "";
  26.     scanf("%[^\n]", str);
  27.     printf("%d", atol(str));
  28.     return 0;
  29. }

元宇宙A+B

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. const static char decToMeta[37] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  5. static char c[100] = "", a[100] = "", b[100] = "";
  6. static int C[100] = {0}, A[100] = {0}, B[100] = {0};
  7.  
  8. int metaToDec(char m) {
  9.     if ('0' <= m && m <= '9') return m - '0';
  10.     return m - 'A' + 10;
  11. }
  12.  
  13. void add(void) {
  14.     int lenA = strlen(a), lenB = strlen(b);
  15.     for (int i = 0; i < lenA; ++i) A[i] = metaToDec(a[lenA - i - 1]);
  16.     for (int i = 0; i < lenB; ++i) B[i] = metaToDec(b[lenB - i - 1]);
  17.  
  18.     int carry = 0;
  19.     int lenC = lenA > lenB ? lenA : lenB;
  20.     for (int i = 0; i < lenC; ++i) {
  21.         C[i] = A[i] + B[i] + carry;
  22.         carry = C[i] / 36;
  23.         C[i] %= 36;
  24.     }
  25.     if (carry != 0) {
  26.         C[lenC] = carry;
  27.         ++lenC;
  28.     }
  29.  
  30.     for (int i = lenC - 1; i >= 0; --i) c[i] = decToMeta[C[lenC - i - 1]];
  31.     c[lenC] = '\0';
  32. }
  33.  
  34. int main() {
  35.     scanf("%s %s", a, b);
  36.     add();
  37.     puts(c);
  38.     return 0;
  39. }

字符串切片

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. int t, n, len;
  5. char str[1000];
  6.  
  7. void strSlice(int begin, int end, int step) {
  8.     char slice[1000] = "";
  9.     int pos = 0;
  10.     if (begin < 0) begin += len;
  11.     if (end < 0) end += len;
  12.     if (end >= begin && step > 0) {
  13.         for (int i = begin; i < end; i += step) {
  14.             slice[pos] = str[i];
  15.             ++pos;
  16.         }
  17.     } else if (end < begin && step < 0) {
  18.         for (int i = begin; i > end; i += step) {
  19.             slice[pos] = str[i];
  20.             ++pos;
  21.         }
  22.     }
  23.     puts(slice);
  24. }
  25.  
  26. void mode(void) {
  27.     len = strlen(str);
  28.     int begin, end, step;
  29.     switch (n) {
  30.         case 3: {
  31.             scanf("%d %d %d", &begin, &end, &step);
  32.             break;
  33.         }
  34.         case 2: {
  35.             scanf("%d %d", &begin, &end);
  36.             step = 1;
  37.             break;
  38.         }
  39.         case 1: {
  40.             scanf("%d", &begin);
  41.             end = len, step = 1;
  42.             break;
  43.         }
  44.     }
  45.     strSlice(begin, end, step);
  46. }
  47.  
  48. int main() {
  49.     scanf("%[^\n] %d", str, &t);
  50.     for (int i = 0; i < t; ++i) {
  51.         scanf("%d", &n);
  52.         mode();
  53.     }
  54.     return 0;
  55. }

分离字符串

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. void split(char *str, char *sep) {
  5.     while (*str) {
  6.         char *flag = strstr(str, sep);
  7.         if (flag == NULL) break;
  8.         char sub[101] = "";
  9.         int len = flag - str;
  10.         memcpy(sub, str, len);
  11.         puts(sub);
  12.         memmove(str - len - strlen(sep), str, strlen(str) + 1);
  13.     }
  14.     puts(str);
  15. }
  16.  
  17. int main() {
  18.     char str[2000] = "", sep[2000] = "";
  19.     scanf("%[^\n] %[^\n]", str, sep);
  20.     split(str, sep);
  21.     return 0;
  22. }

Kids A+B

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. char ans[30] = "";
  5.  
  6. int strToNum(char *str) {
  7.     if(strstr(str, "zero")) return 0;
  8.     if(strstr(str, "ten")) return 10;
  9.     if(strstr(str, "eleven")) return 11;
  10.     if(strstr(str, "twelve")) return 12;
  11.     if(strstr(str, "thirteen")) return 13;
  12.     if(strstr(str, "fourteen")) return 14;
  13.     if(strstr(str, "fifteen")) return 15;
  14.     if(strstr(str, "sixteen")) return 16;
  15.     if(strstr(str, "seventeen")) return 17;
  16.     if(strstr(str, "eighteen")) return 18;
  17.     if(strstr(str, "nineteen")) return 19;
  18.  
  19.     int unit = 0, decade = 0;
  20.     if(strstr(str, "one")) unit = 1;
  21.     if(strstr(str, "two")) unit = 2;
  22.     if(strstr(str, "three")) unit = 3;
  23.     if(strstr(str, "four")) unit = 4;
  24.     if(strstr(str, "five")) unit = 5;
  25.     if(strstr(str, "six")) unit = 6;
  26.     if(strstr(str, "seven")) unit = 7;
  27.     if(strstr(str, "eight")) unit = 8;
  28.     if(strstr(str, "nine")) unit = 9;
  29.     if(strstr(str, "twenty")) decade = 20;
  30.     if(strstr(str, "thirty")) decade = 30;
  31.     if(strstr(str, "forty")) decade = 40;
  32.     if(strstr(str, "fifty")) decade = 50;
  33.     if(strstr(str, "sixty")) decade = 60;
  34.     if(strstr(str, "seventy")) decade = 70;
  35.     if(strstr(str, "eighty")) decade = 80;
  36.     if(strstr(str, "ninety")) decade = 90;
  37.     return unit + decade;
  38. }
  39.  
  40. void numToStr(int n) {
  41.     switch (n) {
  42.         case 0: {
  43.             strcpy(ans, "zero");
  44.             char *p = ans;
  45.             return;
  46.         }
  47.         case 11: {
  48.             strcpy(ans, "eleven");
  49.             char *p = ans;
  50.             return;
  51.         }
  52.         case 12: {
  53.             strcpy(ans, "twelve");
  54.             char *p = ans;
  55.             return;
  56.         }
  57.         case 13: {
  58.             strcpy(ans, "thirteen");
  59.             char *p = ans;
  60.             return;
  61.         }
  62.         case 14: {
  63.             strcpy(ans, "fourteen");
  64.             char *p = ans;
  65.             return;
  66.         }
  67.         case 15: {
  68.             strcpy(ans, "fifteen");
  69.             char *p = ans;
  70.             return;
  71.         }
  72.         case 16: {
  73.             strcpy(ans, "sixteen");
  74.             char *p = ans;
  75.             return;
  76.         }
  77.         case 17: {
  78.             strcpy(ans, "seventeen");
  79.             char *p = ans;
  80.             return;
  81.         }
  82.         case 18: {
  83.             strcpy(ans, "eighteen");
  84.             char *p = ans;
  85.             return;
  86.         }
  87.         case 19: {
  88.             strcpy(ans, "nineteen");
  89.             char *p = ans;
  90.             return;
  91.         }
  92.         default:
  93.             break;
  94.     }
  95.  
  96.     int decade = (n / 10) % 10, unit = n % 10;
  97.     switch (decade) {
  98.         case 2: {
  99.             strcpy(ans, "twenty");
  100.             break;
  101.         }
  102.         case 3: {
  103.             strcpy(ans, "thirty");
  104.             break;
  105.         }
  106.         case 4: {
  107.             strcpy(ans, "forty");
  108.             break;
  109.         }
  110.         case 5: {
  111.             strcpy(ans, "fifty");
  112.             break;
  113.         }
  114.         case 6: {
  115.             strcpy(ans, "sixty");
  116.             break;
  117.         }
  118.         case 7: {
  119.             strcpy(ans, "seventy");
  120.             break;
  121.         }
  122.         case 8: {
  123.             strcpy(ans, "eighty");
  124.             break;
  125.         }
  126.         case 9: {
  127.             strcpy(ans, "ninety");
  128.             break;
  129.         }
  130.         default: {
  131.             break;
  132.         }
  133.     }
  134.  
  135.     if (decade && unit) strcat(ans, "-");
  136.  
  137.     switch (unit) {
  138.         case 1: {
  139.             strcat(ans, "one");
  140.             break;
  141.         }
  142.         case 2: {
  143.             strcat(ans, "two");
  144.             break;
  145.         }
  146.         case 3: {
  147.             strcat(ans, "three");
  148.             break;
  149.         }
  150.         case 4: {
  151.             strcat(ans, "four");
  152.             break;
  153.         }
  154.         case 5: {
  155.             strcat(ans, "five");
  156.             break;
  157.         }
  158.         case 6: {
  159.             strcat(ans, "six");
  160.             break;
  161.         }
  162.         case 7: {
  163.             strcat(ans, "seven");
  164.             break;
  165.         }
  166.         case 8: {
  167.             strcat(ans, "eight");
  168.             break;
  169.         }
  170.         case 9: {
  171.             strcat(ans, "nine");
  172.             break;
  173.         }
  174.         default: {
  175.             break;
  176.         }
  177.     }
  178. }
  179.  
  180. int main() {
  181.     char a[30] = "", b[30] = "";
  182.     scanf("%s %s", a, b);
  183.  
  184.     numToStr(strToNum(a) + strToNum(b));
  185.     puts(ans);
  186.     return 0;
  187. }

61-70

时钟A-B

  1. #include <stdio.h>
  2. #include <time.h>
  3.  
  4. int main(){
  5. 	struct tm begin = {0}, end = {0};
  6. 	scanf("%d %d %d", &begin.tm_year, &begin.tm_mon, &begin.tm_mday);
  7. 	scanf("%d %d %d", &end.tm_year, &end.tm_mon, &end.tm_mday);
  8.  
  9. 	begin.tm_year -= 1900, begin.tm_mon -= 1;
  10. 	end.tm_year -= 1900, end.tm_mon -= 1;
  11.  
  12. 	time_t tmBegin =  mktime(&begin);
  13. 	time_t tmEnd = mktime(&end);
  14. 	printf("%.6lf", difftime(tmBegin, tmEnd));
  15. 	return 0;
  16. }

加密字串

  1. #include <stdio.h>
  2.  
  3. static int freq[26] = {0};
  4.  
  5. int main() {
  6. 	char plain[8000] = "";
  7. 	int x;
  8. 	scanf("%s %d", plain, &x);
  9. 	for (int i = 0; plain[i]; ++i) ++freq[plain[i] - 'a'];
  10. 	char cipher[8000] = "";
  11. 	for (int i = 0; plain[i]; ++i) {
  12. 		if (freq[plain[i] - 'a'] & 1)
  13. 			cipher[i] = (char) (((plain[i] - 'a' - x) % 26 + 26) % 26 + 'a');
  14. 		else
  15. 			cipher[i] = (char) ((plain[i] - 'a' + x) % 26 + 'a');
  16. 	}
  17. 	puts(cipher);
  18. 	return 0;
  19. }

Arduino显示

  1. #include <stdio.h>
  2.  
  3. static const int digit[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  4.  
  5. int getUnit(int num) {
  6. 	int cnt = 0;
  7. 	do {
  8. 		cnt += digit[num % 10];
  9. 		num /= 10;
  10. 	} while (num);
  11. 	return cnt;
  12. }
  13.  
  14. int main() {
  15. 	int n;
  16. 	scanf("%d", &n);
  17. 	n -= 4;
  18. 	if (n <= 0) printf("0");
  19. 	else {
  20. 		int cnt = 0;
  21. 		for (int i = 0; i <= 1111; ++i) {
  22. 			for (int j = 0; j <= 1111; ++j) {
  23. 				if (getUnit(i) + getUnit(j) + getUnit(i + j) == n) ++cnt;
  24. 			}
  25. 		}
  26. 		printf("%d", cnt);
  27. 	}
  28. 	return 0;
  29. }

有效表达式

  1. #include <stdio.h>
  2.  
  3. int main() {
  4. 	long long n;
  5. 	scanf("%lld", &n);
  6. 	long long cnt = 1;
  7. 	for (long long i = n + 2; i <= 2 * n; ++i) cnt *= i;
  8. 	for (long long i = 1; i <= n; ++i) cnt /= i;
  9. 	printf("%lld", cnt);
  10. 	return 0;
  11. }

长安

  1. #include <stdio.h>
  2.  
  3. int bx, by, px, py, cnt;
  4.  
  5. void dfs(int x, int y) {
  6. 	if ((x == px && y == py) || x > bx || y > by) return;
  7. 	if (x == bx && y == by) {
  8. 		++cnt;
  9. 		return;
  10. 	}
  11. 	dfs(x + 1, y);
  12. 	dfs(x, y + 1);
  13. }
  14.  
  15. int main() {
  16. 	while (1) {
  17. 		fflush(stdin);
  18. 		scanf("%d %d %d %d", &bx, &by, &px, &py);
  19. 		if (bx <= 0 || by <= 0 || px <= 0 || py <= 0) break;
  20. 		cnt = 0;
  21. 		dfs(1, 1);
  22. 		printf("%d\n", cnt);
  23. 	}
  24. 	return 0;
  25. }

GPS通信协议（CPP）

感谢 Sekiro_2 提供的代码~

  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. string out[100];
  6. int k=0;
  7.  
  8. int check(string str){
  9.     int i,result;
  10.  
  11.     for(result=str[1],i=2;str[i]!='*';i++)
  12.     {
  13.         result^=str[i];
  14.     }
  15.     return result;
  16. }
  17.  
  18. int convert(string str){
  19.     int res=0;
  20.     res=stoi(str,0,16);
  21.     return res;
  22. }
  23.  
  24.  
  25. void convert_BeingTime(string utcTime){
  26.     int  hour=stoi(utcTime.substr(0,2));
  27.     int  B_hour=(hour+8)%24;
  28.     if(B_hour/10==0)
  29.         out[k++]="0"+to_string(B_hour)+":"+utcTime.substr(2,2)+":"+utcTime.substr(4,2);
  30.     else
  31.         out[k++]=to_string(B_hour)+":"+utcTime.substr(2,2)+":"+utcTime.substr(4,2);
  32.  
  33. }
  34.  
  35. int main(){
  36.     string str;
  37.     while(cin>>str){
  38.         if(str=="END") break;
  39.         if(str.compare(0,6,"$GPRMC")==0){
  40.             size_t asteriskPos = str.find('*');
  41.             if(asteriskPos!=string::npos){
  42.                 int checksum=check(str);
  43.                 
  44.                 int senchecksum=convert(str.substr(asteriskPos + 1, 2));
  45.                 if(checksum!=senchecksum) {
  46.                     out[k++]="error";
  47.           
  48.                 }
  49.                 else{
  50.                     string utcTime = str.substr(7, 6);
  51.      
  52.                     convert_BeingTime(utcTime);
  53.                 }
  54.             }
  55.         }
  56.  
  57.     }
  58.     for(int i=0;i<k;i++){
  59.         cout<<out[i]<<endl;
  60.     }
  61. }

三元搜索

  1. #include <stdio.h>
  2.  
  3. int terSearch(int arr[], int n, int k) {
  4. 	int left = 0, right = n - 1, mid1 = (n - 1) / 3, mid2 = n - mid1;
  5. 	while(mid1 != mid2) {
  6. 		if (k > arr[right] || k < arr[left]) return -1;
  7. 		if (k == arr[mid1]) return mid1;
  8. 		if (k == arr[mid2]) return mid2;
  9. 		if (mid1 == mid2) break;
  10. 		if (k < arr[mid1]) right = mid1 - 1;
  11. 		else if (k > arr[mid2]) left = mid2 + 1;
  12. 		else left = mid1 + 1, right = mid2 - 1;
  13. 		mid1 = left + (right - left) / 3, mid2 = right - (right - left) / 3;
  14. 	}
  15. 	return -1;
  16. }
  17.  
  18. int main() {
  19. 	int n, k;
  20. 	scanf("%d", &n);
  21. 	int arr[n];
  22. 	for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  23. 	scanf("%d", &k);
  24. 	printf("%d in [%d]", k, terSearch(arr, n, k));
  25. 	return 0;
  26. }

DNA双螺旋结构

  1. #include <stdio.h>
  2.  
  3. void putsDna1(){
  4. 	printf("   AT   \n");
  5. 	printf("  T--A  \n");
  6. 	printf(" A----T \n");
  7. 	printf("T------A\n");
  8. 	printf("T------A\n");
  9. 	printf(" G----C \n");
  10. 	printf("  T--A  \n");
  11. 	printf("   GC   \n");
  12. }
  13.  
  14. void putsDna2(){
  15. 	printf("   CG   \n");
  16. 	printf("  C--G  \n");
  17. 	printf(" A----T \n");
  18. 	printf("A------T\n");
  19. 	printf("T------A\n");
  20. 	printf(" A----T \n");
  21. 	printf("  A--T  \n");
  22. 	printf("   GC   \n");
  23. }
  24.  
  25. void putsDna3(){
  26. 	printf("   AT   \n");
  27. 	printf("  C--G  \n");
  28. 	printf(" T----A \n");
  29. 	printf("C------G\n");
  30. 	printf("C------G\n");
  31. 	printf(" T----A \n");
  32. 	printf("  G--C  \n");
  33. 	printf("   AT   \n");
  34. }
  35.  
  36. int main() {
  37. 	int n;
  38. 	scanf("%d", &n);
  39. 	for (int i = 1; i <= n/2; ++i) {
  40. 		if (i % 3 == 1) putsDna1();
  41. 		else if (i % 3 == 2) putsDna2();
  42. 		else putsDna3();
  43. 	}
  44. 	return 0;
  45. }

PID控制

  1. #include <stdio.h>
  2.  
  3. typedef struct PIDController {
  4. 	double Kp, Ki, Kd;
  5. 	double preError, integral;
  6. } PIDData;
  7.  
  8. double PIDCalculate(PIDData *pid, double setPoint, double measuredValue) {
  9. 	double error = setPoint - measuredValue;
  10. 	pid->integral += error;
  11. 	double differential = error - pid->preError;
  12. 	double output = pid->Kp * error + pid->Ki * pid->integral + pid->Kd * differential;
  13. 	pid->preError = error;
  14. 	return output;
  15. }
  16.  
  17. int main() {
  18. 	double setPoint, measuredValue;
  19. 	int time;
  20. 	PIDData pid = {0};
  21. 	scanf("%lf %lf %lf", &pid.Kp, &pid.Ki, &pid.Kd);
  22. 	scanf("%lf %lf %d", &setPoint, &measuredValue, &time);
  23. 	for (int i = 1; i <= time; ++i) {
  24. 		double output = PIDCalculate(&pid, setPoint, measuredValue);
  25. 		measuredValue += output;
  26. 		printf("%d %.6lf\n", i, measuredValue);
  27. 	}
  28. 	return 0;
  29. }

循环排序

  1. #include <stdio.h>
  2.  
  3. void swap(int *a, int *b) {
  4. 	int tmp = *a;
  5. 	*a = *b, *b = tmp;
  6. }
  7.  
  8. void cycleSort(int arr[], int n) {
  9. 	for (int i = 0; i < n - 1; ++i) {
  10. 		int item = arr[i], pos = i;
  11. 		for (int j = i + 1; j < n; ++j) if (arr[j] < item) ++pos;
  12. 		if (pos == i) continue;
  13.  
  14. 		swap(&arr[pos], &item);
  15. 		while(pos != i) {
  16. 			pos = i;
  17. 			for (int j = i + 1; j < n; ++j) if (arr[j] < item) ++pos;
  18. 			while (item == arr[pos]) ++pos;
  19. 			swap(&arr[pos], &item);
  20. 		}
  21. 	}
  22. }
  23.  
  24. int main() {
  25. 	int n;
  26. 	scanf("%d", &n);
  27. 	int arr[n];
  28. 	for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  29. 	cycleSort(arr, n);
  30. 	for (int i = 0; i < n; ++i) printf("%d ", arr[i]);
  31. 	return 0;
  32. }

71-80

卫星定位（WA）

热能计算

  1. #include <stdio.h>
  2.  
  3. int main() {
  4. 	int Ti, Tf, cL, cV;
  5. 	double mL, mV;
  6. 	scanf("%d %d %lf %d %lf %d", &Ti, &Tf, &mL, &cL, &mV, &cV);
  7. 	double QL = cL * mL * (Tf - Ti);
  8. 	double QV = cV * mV * (Tf - Ti);
  9. 	double Q = QL + QV;
  10. 	printf("%.2lfkJ,%.2lf%%,%.2lf%%\n", Q / 1000, QV / Q, QL / Q);
  11. 	return 0;
  12. }
  13.  
  14. /*
  15. 20 80
  16. 0.250 4186
  17. 0.500 900
  18. */

几何约束

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3.  
  4. int maxi(int a, int b) {
  5. 	return a > b ? a : b;
  6. }
  7.  
  8. int mini(int a, int b) {
  9. 	return a < b ? a : b;
  10. }
  11.  
  12. int cross(int x1, int y1, int x2, int y2) {
  13. 	return x1 * y2 - y1 * x2;
  14. }
  15.  
  16. bool insert(int line1[4], int line2[4]) {
  17. 	if (maxi(line2[0], line2[2]) < mini(line1[0], line1[2]) ||
  18. 	maxi(line1[0],line1[2]) < mini(line2[0], line2[2]) ||
  19. 	maxi(line2[1], line2[3]) < mini(line1[1], line1[3]) ||
  20. 	 maxi(line1[1], line2[3]) < mini(line2[1],line2[3])) return false;
  21. 	if (cross(line1[2] - line1[0], line1[3] - line1[1], line2[0] - line1[0], line2[1] - line1[1]) *
  22. 		cross(line1[2] - line1[0], line1[3] - line1[1], line2[2] - line1[0], line2[3] - line1[1]) > 0 ||
  23. 		cross(line2[2] - line2[0], line2[3] - line2[1], line1[0] - line2[0], line1[1] - line2[1]) *
  24. 			cross(line2[2] - line2[0], line2[3] - line2[1], line1[2] - line2[0], line1[3] - line2[1]) > 0) return false;
  25. 	return true;
  26. }
  27.  
  28. int main() {
  29. 	int n;
  30. 	scanf("%d", &n);
  31. 	int line[n][4];
  32. 	for (int i = 0; i < n; ++i)
  33. 		for (int j = 0; j < 4; ++j) scanf("%d", &line[i][j]);
  34.  
  35. 	int cnt = 0;
  36. 	for (int i = 0; i < n; ++i) {
  37. 		for (int j = i + 1; j < n; ++j) {
  38. 			if (insert(line[i], line[j])) {
  39. 				printf("X: #%d #%d\n", i + 1, j + 1);
  40. 				++cnt;
  41. 			}
  42. 		}
  43. 	}
  44. 	printf("n=%d\n", cnt);
  45. 	return 0;
  46. }
  47.  
  48. /*
  49. 5
  50. 1 5 4 5
  51. 2 5 10 1
  52. 3 2 10 3
  53. 6 4 9 4
  54. 7 1 8 1
  55. */

日出日落时间（WA）

中位数

  1. #include <stdio.h>
  2.  
  3. int arr[1000];
  4.  
  5. double mid(int n) {
  6. 	if (n & 1) return arr[n / 2];
  7. 	return (arr[n / 2] + arr[n / 2 - 1]) / 2.0f;
  8. }
  9.  
  10. int main() {
  11. 	int flag, cnt = 0;
  12. 	while(1) {
  13. 		scanf("%d", &flag);
  14. 		if (flag == -1) break;
  15. 		while (1) {
  16. 			if (flag == 0) {
  17. 				for (int i = 0; i < cnt; ++i) printf("%d ", arr[i]);
  18. 				printf("%.6lf\n", mid(cnt));
  19. 				break;
  20. 			}
  21. 			arr[cnt++] = flag;
  22. 			scanf("%d", &flag);
  23. 		}
  24. 	}
  25. 	return 0;
  26. }

原子计数（CPP）

感谢 Cubeist 提供的代码~

  1. #include <iostream>
  2. #include <map>
  3. using namespace std;
  4.  
  5. string s;
  6.  
  7. int getnum(int x)
  8. {
  9.     int res = 0;
  10.     for (; s[x]>='0' && s[x]<='9' && s[x]; x ++)
  11.         res = res * 10 + s[x] - '0';
  12.     return res + !res;
  13. }
  14.  
  15. int main()
  16. {
  17.     getline(cin, s);
  18.     map<string, int> mp;
  19.     for (int i = 0; s[i]; i ++)
  20.     {
  21.         if (!(s[i]>='A' && s[i]<='Z')) continue;
  22.         string ele = "";
  23.         ele += s[i];
  24.         
  25.         if (s[i+1]>='a' && s[i+1]<='z')
  26.         {
  27.             ele += s[i+1];
  28.             mp[ele] += getnum(i+2);
  29.         }
  30.         else 
  31.             mp[ele] += getnum(i+1);
  32.     }
  33.  
  34.     for (auto& p : mp)
  35.         cout << p.first << " " << p.second << endl;
  36.         
  37.     return 0;
  38. }

成绩单

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <stdbool.h>
  4.  
  5. typedef struct tag {
  6. 	long long id;
  7. 	char name[31];
  8. 	int score;
  9. } ST;
  10.  
  11. void load(ST* arr[], int n) {
  12. 	for (int i = 0; i < n; ++i) {
  13. 		arr[i] = (ST*) malloc(sizeof(ST));
  14. 		scanf("%lld %s %d", &arr[i]->id, arr[i]->name, &arr[i]->score);
  15. 	}
  16. }
  17.  
  18. void sort(ST* arr[], int n) {
  19. 	ST* tmp = NULL;
  20. 	for (int i = 0; i < n; ++i) {
  21. 		bool isSwapped = false;
  22. 		for (int j = 0; j < n - 1 - i; ++j)
  23. 			if ((arr[j]->score < arr[j + 1]->score) ||
  24. 				(arr[j]->score == arr[j + 1]->score && arr[j]->id > arr[j + 1]->id))
  25. 				tmp = arr[j], arr[j] = arr[j + 1], arr[j + 1] = tmp, isSwapped = true;
  26. 		if (!isSwapped) break;
  27. 	}
  28. }
  29.  
  30. void traverse(ST* arr[], int n) {
  31. 	for (int i = 0; i < n; ++i)
  32. 		printf("%lld %s %d\n", arr[i]->id, arr[i]->name, arr[i]->score);
  33. }
  34.  
  35. int main() {
  36. 	int n;
  37. 	scanf("%d", &n);
  38. 	ST* grade[n];
  39. 	load(grade, n);
  40. 	sort(grade, n);
  41. 	traverse(grade, n);
  42. 	return 0;
  43. }
  44.  
  45. /*
  46. 6
  47. 2001900001 Jerry 88
  48. 2001900005 Tom 92
  49. 2001900006 Milla 85
  50. 2001900002 Alice 80
  51. 2001900003 Mickey 85
  52. 2001900004 Aladdin 83
  53. */

水下声学定位

  1. #include<stdio.h>
  2. #include<math.h>
  3.  
  4. int main()
  5. {
  6. 	const double pi = 3.1415926f;
  7. 	double a, b, c, d, diag;
  8. 	scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &d, &diag);
  9. 	double p = (a + b + diag) / 2, q = (c + d + diag);
  10. 	double s = sqrt(p * (p - a) * (p - b) * (p - diag)) + sqrt(q * (q - c) * (q - d) * (q - diag));
  11. 	double angle = atan((4 * s) / (pow(b, 2) + pow(d, 2) - pow(a, 2) - pow(c, 2)));
  12. 	angle *= 180 / pi;
  13. 	printf("%.6lf %.1lf", s, angle);
  14. }

火箭发射模拟（CPP）

感谢 De1ta_Zer0 提供的代码~

  1. #include <iostream>
  2. #include <iomanip>
  3.  
  4. const double timeStep = 0.1;
  5.  
  6. int main()
  7. {
  8.     double totalMass, rocketMass, burnTime, cE, g;
  9.     std::cin >> totalMass >> rocketMass >> burnTime >> cE >> g;
  10.     double propellantMass = totalMass - rocketMass;
  11.     double massFlow = propellantMass / burnTime;
  12.     double T = massFlow * cE;
  13.     double altitude = 0, V = 0;
  14.     double timeLeft = burnTime + timeStep;
  15.     while(timeLeft >= 0)
  16.     {
  17.         double a = T / totalMass;
  18.         double deltaV = a * timeStep;
  19.         double deltaAltitude = V * timeStep;
  20.         double deltaM = massFlow * timeStep;
  21.         V += deltaV;
  22.         altitude += deltaAltitude;
  23.         totalMass -= deltaM;
  24.         timeLeft -= timeStep;
  25.     }
  26.     std::cout << std::fixed << std::setprecision(3) << altitude / 1000 << "km" << std::endl;
  27.     return 0;
  28. }

晶体结构（CPP）

感谢 De1ta_Zer0 提供的代码~

  1. #include <iostream>
  2. #include <string>
  3. #include <iomanip>
  4. #include <cmath>
  5.  
  6. struct Atom
  7. {
  8.     std::string name;
  9.     double mass;
  10.     double APF;
  11.     double r;
  12. };
  13.  
  14. Atom elemList[] =
  15. {
  16.     { "Po",   208.998, 0.52360, 1.68 },
  17.     { "Li",     6.941, 0.68017, 1.52 },
  18.     { "Na", 22.989770, 0.68017, 1.86 },
  19.     { "Cr",   51.9961, 0.68017, 1.28 },
  20.     { "Mn", 54.938049, 0.68017, 1.27 },
  21.     { "Fe",    55.845, 0.68017, 1.26 },
  22.     { "Mo",     95.94, 0.68017, 1.39 },
  23.     { "Ta",  180.9749, 0.68017, 1.46 },
  24.     { "Al", 26.981538, 0.74048, 1.43 },
  25.     { "Ca",    40.078, 0.74048, 1.97 },
  26.     { "Ni",   58.6934, 0.74048, 1.24 },
  27.     { "Cu",    63.546, 0.74048, 1.28 },
  28.     { "Ge",     72.64, 0.74048, 1.22 },
  29.     { "Ag",  107.8682, 0.74048, 1.44 },
  30.     { "Pt",   195.078, 0.74048, 1.39 },
  31.     { "Au", 196.96655, 0.74048, 1.44 },
  32.     { "Pb",     207.2, 0.74048, 1.75 }
  33. };
  34.  
  35. template <int N> double pow(double a) { return a * pow<N - 1>(a); }
  36.  
  37. template <> double pow<0>(double a) { return 1; }
  38.  
  39. int main()
  40. {
  41.  
  42.     int n;
  43.     std::cin >> n;
  44.     for (int i = 0; i < n; ++i)
  45.     {
  46.         std::string in;
  47.         std::cin >> in;
  48.         for(const auto& e : elemList)
  49.         {
  50.             if(e.name == in)
  51.             {
  52.                 double V = 4.0 * M_PI * pow<3>(e.r) / 3.0;
  53.                 std::cout << std::fixed << std::setprecision(2) << 10.0 * e.mass * e.APF / 6.022 / V << std::endl;
  54.                 break;
  55.             }
  56.         }
  57.     }
  58.     return 0;
  59. }

81-90

上楼梯

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. int main() {
  5. 	int n, m;
  6. 	scanf("%d %d", &n, &m);
  7. 	long long dp[n + 1];
  8. 	memset(dp, -1, (n + 1) * sizeof(long long));
  9. 	for (int i = 0; i < m; ++i) {
  10. 		int tmp;
  11. 		scanf("%d", &tmp);
  12. 		dp[tmp] = 0;
  13. 	}
  14.  
  15. 	dp[1] = dp[1] ? 1 : 0;
  16. 	dp[2] = dp[2] ? (dp[1] ? 2 : 1) : 0;
  17. 	for (int i = 3; i <= n; ++i)
  18. 		if(dp[i]) dp[i] = (dp[i - 1] + dp [i - 2]) % (long long) (1e9 + 7);
  19. 	printf("%lld\n", dp[n]);
  20. 	return 0;
  21. }

绝对差

  1. #include <stdio.h>
  2. #include <limits.h>
  3. #include <time.h>
  4. #include <stdlib.h>
  5.  
  6. void quickSort(int arr[], int left, int right) {
  7. 	if (left >= right) return;
  8. 	srand(time(NULL));
  9. 	int idx = rand() % (left - right) + left;
  10. 	int flag = arr[idx], head = left - 1, tail = right + 1;
  11. 	while (head < tail) {
  12. 		do head++; while(arr[head] < flag);
  13. 		do tail--; while(arr[tail] > flag);
  14. 		if (head < tail) {
  15. 			int tmp = arr[head];
  16. 			arr[head] = arr[tail];
  17. 			arr[tail] = tmp;
  18. 		}
  19. 	}
  20. 	quickSort(arr, left, tail);
  21. 	quickSort(arr, tail + 1, right);
  22. }
  23.  
  24. int minAbsSub(int arr[], int n) {
  25. 	int min = INT_MAX;
  26. 	for (int i = 0; i < n - 1; ++i) {
  27. 		int tmp = arr[i + 1] - arr[i];
  28. 		if (tmp < min) min = tmp;
  29. 	}
  30. 	return min;
  31. }
  32.  
  33. int main() {
  34. 	int n;
  35. 	scanf("%d", &n);
  36. 	int arr[n];
  37. 	for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  38.  
  39. 	quickSort(arr, 0, n - 1);
  40. 	printf("%d", minAbsSub(arr, n));
  41. 	return 0;
  42. }

挑选

  1. #include <stdio.h>
  2. #include <limits.h>
  3. #include <stdlib.h>
  4. #include <time.h>
  5.  
  6. void quickSort(long long arr[], long long left, long long right) {
  7. 	if (left >= right) return;
  8. 	srand(time(NULL));
  9. 	long long idx = rand() % (left - right) + left;
  10. 	long long flag = arr[idx], head = left - 1, tail = right + 1;
  11. 	while (head < tail) {
  12. 		do head++; while(arr[head] < flag);
  13. 		do tail--; while(arr[tail] > flag);
  14. 		if (head < tail) {
  15. 			long long tmp = arr[head];
  16. 			arr[head] = arr[tail];
  17. 			arr[tail] = tmp;
  18. 		}
  19. 	}
  20. 	quickSort(arr, left, tail);
  21. 	quickSort(arr, tail + 1, right);
  22. }
  23.  
  24. long long maxi(long long arr[], long long n) {
  25. 	long long max = LLONG_MIN;
  26. 	for (long long i = 0; i < n; ++i)
  27. 		if (max < arr[i]) max = arr[i];
  28. 	return max;
  29. }
  30.  
  31. long long sumMax(long long arr[], long long n) {
  32. 	long long dp[n];
  33. 	dp[0] = arr[0];
  34. 	for (long long i = 1; i < n; ++i) {
  35. 		if (arr[i] == arr[i - 1])
  36. 			dp[i] = dp[i - 1] > (arr[i] + dp[i - 1]) ? dp[i - 1] : (arr[i] + dp[i - 1]);
  37. 		else {
  38. 			long long j = i - 1;
  39. 			while (j >= 0 && arr[j] >= arr[i] - 1) --j;
  40. 			dp[i] = arr[i] + (j >= 0 ? dp[j] : 0);
  41. 		}
  42. 	}
  43. 	return maxi(dp, n);
  44. }
  45.  
  46. int main() {
  47. 	long long n;
  48. 	scanf("%lld", &n);
  49. 	long long arr[n];
  50. 	for (long long i = 0; i < n; ++i) scanf("%lld", &arr[i]);
  51.  
  52. 	quickSort(arr, 0, n - 1);
  53. 	printf("%lld\n", sumMax(arr, n));
  54. 	return 0;
  55. }

三角形

  1. #include <stdio.h>
  2. #include <time.h>
  3. #include <stdlib.h>
  4.  
  5. void quickSort(int arr[], int left, int right) {
  6. 	if (left >= right) return;
  7. 	srand(time(NULL));
  8. 	int idx = rand() % (left - right) + left;
  9. 	int flag = arr[idx], head = left - 1, tail = right + 1;
  10. 	while (head < tail) {
  11. 		do head++; while(arr[head] < flag);
  12. 		do tail--; while(arr[tail] > flag);
  13. 		if (head < tail) {
  14. 			int tmp = arr[head];
  15. 			arr[head] = arr[tail];
  16. 			arr[tail] = tmp;
  17. 		}
  18. 	}
  19. 	quickSort(arr, left, tail);
  20. 	quickSort(arr, tail + 1, right);
  21. }
  22.  
  23. void findTri(int arr[], int n) {
  24. 	for (int i = n - 1; i > 1; --i)
  25. 		if (arr[i] < arr[i - 1] + arr[i - 2]) {
  26. 			printf("%d %d %d\n", arr[i - 2], arr[i - 1], arr[i]);
  27. 			return;
  28. 		}
  29. 	printf("-1\n");
  30. }
  31.  
  32. int main() {
  33. 	int n;
  34. 	scanf("%d", &n);
  35. 	int arr[n];
  36. 	for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  37.  
  38. 	quickSort(arr, 0, n - 1);
  39. 	findTri(arr, n);
  40. 	return 0;
  41. }

子数组最大和

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. int maxSum(int arr[], int n) {
  5. 	int dp[n];
  6. 	memset(dp, 0, n * sizeof(int));
  7. 	int maxi = arr[0];
  8. 	dp[0] = arr[0];
  9. 	for (int i = 1; i < n; ++i) {
  10. 		dp[i] = arr[i] > (dp[i - 1] + arr[i]) ? arr[i] : (dp[i - 1] + arr[i]);
  11. 		maxi = dp[i] > maxi ? dp[i] : maxi;
  12. 	}
  13. 	return maxi;
  14. }
  15.  
  16. int main() {
  17. 	int n;
  18. 	scanf("%d", &n);
  19. 	int arr[n];
  20. 	for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  21. 	printf("%d\n", maxSum(arr, n));
  22. 	return 0;
  23. }

和字符串

  1. #include <stdio.h>
  2. #include <string.h>
  3. #include <stdbool.h>
  4.  
  5. long long substrToNum(char str[], int pos, int len) {
  6. 	long long num = 0;
  7. 	for (int i = 0; i < len; ++i)
  8. 		num = num * 10 + str[pos + i] - '0';
  9. 	return num;
  10. }
  11.  
  12. long long getLen(long long n) {
  13. 	int cnt = 0;
  14. 	do ++cnt, n /= 10; while(n);
  15. 	return cnt;
  16. }
  17.  
  18. bool backTracking(char str[], int strLen, int begin, int len1, int len2) {
  19. 	if (begin + len1 + len2 >= strLen) return true;
  20. 	long long num1 = substrToNum(str, begin, len1);
  21. 	long long num2 = substrToNum(str, begin + len1, len2);
  22. 	long long num3 = substrToNum(str, begin + len1 + len2, getLen(num1 + num2));
  23. 	if (num1 + num2 == num3) return backTracking(str, strLen, begin + getLen(num1), getLen(num2), getLen(num3));
  24. 	return false;
  25. }
  26.  
  27. void partition(char str[]) {
  28. 	int strLen = strlen(str);
  29. 	for (int i = 1; i <= strLen / 2; ++i) {
  30. 		if (backTracking (str, strLen, 0, i, i)) {
  31. 			printf("true\n");
  32. 			return;
  33. 		}
  34. 	}
  35. 	printf("false\n");
  36. }
  37.  
  38. int main() {
  39. 	char str[1000] = "";
  40. 	scanf("%s", str);
  41. 	partition(str);
  42. 	return 0;
  43. }

汤包

  1. #include <stdio.h>
  2. #include <time.h>
  3. #include <stdlib.h>
  4.  
  5. typedef struct {
  6. 	int guest;
  7. 	int end;
  8. } Tag;
  9.  
  10. void quickSort(Tag* arr[], int left, int right) {
  11. 	if (left >= right) return;
  12. 	srand(time(NULL));
  13. 	int idx = rand() % (left - right) + left;
  14. 	int flag = arr[idx]->end, head = left - 1, tail = right + 1;
  15. 	while (head < tail) {
  16. 		do head++; while(arr[head]->end < flag);
  17. 		do tail--; while(arr[tail]->end > flag);
  18. 		if (head < tail) {
  19. 			Tag *tmp = arr[head];
  20. 			arr[head] = arr[tail];
  21. 			arr[tail] = tmp;
  22. 		}
  23. 	}
  24. 	quickSort(arr, left, tail);
  25. 	quickSort(arr, tail + 1, right);
  26. }
  27.  
  28. void traverse(Tag *arr[], int n) {
  29. 	for (int i = 0; i < n; ++i)
  30. 		printf("%d ", arr[i]->guest);
  31. }
  32.  
  33. int main() {
  34. 	int n;
  35. 	scanf("%d", &n);
  36. 	Tag *arr[n];
  37. 	for (int i = 0; i < n; ++i) {
  38. 		arr[i] = (Tag*)malloc(sizeof(Tag));
  39. 		arr[i]->guest = i + 1;
  40. 		int begin, duration;
  41. 		scanf("%d %d", &begin, &duration);
  42. 		arr[i]->end = begin + duration;
  43. 	}
  44.  
  45. 	quickSort(arr, 0, n -1);
  46. 	traverse(arr, n);
  47. 	return 0;
  48. }

打字机

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. long long dp[100];
  5. long long seg[100] = {0};
  6.  
  7. long long method(char str[]) {
  8. 	memset(dp, -1, 100 * sizeof(long long));
  9. 	char *iter = str;
  10. 	int part = 0, ans = 1;
  11. 	seg[part] = 1;
  12. 	while (*iter) {
  13. 		if (*iter == 'm' || *iter == 'w') return 0;
  14. 		if (*iter == 'n' && *(iter + 1) == 'n') {
  15. 			int cnt = 1;
  16. 			dp[0] = 1, dp[1] = 2, iter += 2;
  17. 			while(*iter == 'n') {
  18. 				++cnt, ++iter;
  19. 				dp[cnt] = dp[cnt - 1] + dp[cnt - 2];
  20. 			}
  21. 			seg[++part] = dp[cnt];
  22. 		}
  23. 		else if (*iter == 'u' && *(iter + 1) == 'u') {
  24. 			int cnt = 1;
  25. 			dp[0] = 1, dp[1] = 2, iter += 2;
  26. 			while(*iter == 'u') {
  27. 				++cnt, ++iter;
  28. 				dp[cnt] = dp[cnt - 1] + dp[cnt - 2];
  29. 			}
  30. 			seg[++part] = dp[cnt];
  31. 		}
  32. 		else ++iter;
  33. 	}
  34. 	for (int i = 0; seg[i] ; ++i) ans *= seg[i];
  35. 	return ans;
  36. }
  37.  
  38. int main() {
  39. 	char str[1000];
  40. 	scanf("%s", str);
  41. 	printf("%lld", method(str));
  42. 	return 0;
  43. }

游乐园

  1. #include <stdio.h>
  2. #include <stdbool.h>
  3. #include <string.h>
  4.  
  5. int dist[12][12];
  6. int n, m;
  7. bool pass[12];
  8. int ans = -1;
  9.  
  10. bool check(int v, int u) {
  11. 	return !pass[u] && dist[v][u] != 0x3f3f3f3f;
  12. }
  13.  
  14. bool noway(int v) {
  15. 	for (int i = 0; i < n; ++i)
  16. 		if (check(v, i)) return false;
  17. 	return true;
  18. }
  19.  
  20. void backTracking(int v, int l) {
  21. 	if (noway(v)) {
  22. 		ans = ans > l ? ans : l;
  23. 		return;
  24. 	}
  25. 	for (int i = 0; i < n; ++i) {
  26. 		if (check(v, i) && i != v) {
  27. 			pass[i] = true;
  28. 			backTracking(i, l + dist[v][i]);
  29. 			pass[i] = false;
  30. 		}
  31. 	}
  32. }
  33.  
  34. int main() {
  35. 	scanf("%d %d", &n, &m);
  36. 	memset(dist, 0x3f, sizeof(dist));
  37. 	for (int i = 0; i < n; ++i) dist[i][i] = 0;
  38. 	while (m) {
  39. 		int v, u, l;
  40. 		scanf("%d %d %d", &v, &u, &l);
  41. 		v -= 1, u -= 1;
  42. 		dist[v][u] = dist[v][u] < l ? dist[v][u] : l;
  43. 		dist[u][v] = dist[v][u], --m;
  44. 	}
  45.  
  46. 	for (int i = 0; i < n; ++i) {
  47. 		memset(pass, 0, sizeof(pass));
  48. 		pass[i] = true;
  49. 		backTracking(i, 0);
  50. 	}
  51. 	printf("%d\n", ans);
  52. 	return 0;
  53. }

危险的组合（CPP）

感谢 Cubeist 提供的代码~

  1. #include <iostream>
  2. using namespace std;
  3.  
  4. long long w[] = {0, 0, 0, 1, 3, 8, 20, 47, 107, 238, 520, 1121,
  5. 2391, 5056, 10616, 22159, 46023, 95182, 196132, 402873, 825259,
  6. 1686408, 3438828, 6999071, 14221459, 28853662, 58462800,
  7. 118315137, 239186031, 483072832, 974791728};
  8.  
  9. int main()
  10. {
  11.     int n;
  12.     while (cin >> n, n)
  13.     {
  14.         if (n <= 0) exit(0);
  15.         cout << w[n] << endl;
  16.     }
  17.     return 0;
  18. }

91-100 考试模拟

【循环】圆周率 $\pi$

  1. #include <stdio.h>
  2.  
  3. double pi(int n) {
  4. 	double sum = 3.0f;
  5. 	for (int i = 2; i <= n; ++i) {
  6. 		double sgn = i % 2 ? -1.0f : 1.0;
  7. 		sum += sgn * 4 / (2 * i * (2 * i - 1) * (2 * i - 2));
  8. 	}
  9. 	return sum;
  10. }
  11.  
  12. int main() {
  13. 	int n;
  14. 	scanf("%d", &n);
  15. 	printf("%.7lf\n", pi(n));
  16. 	return 0;
  17. }

【选择】马赫数

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. void mach(double v, double T) {
  5. 	double m = (v / 3.6f) / (331.3f * sqrt(1 + T / 273.15f));
  6. 	printf("%.3lf ", m);
  7. 	if (m - 0.8f < 1e-6) printf("subsonic\n");
  8. 	else if (m - 1.2f < 1e-6) printf("transonic\n");
  9. 	else if (m - 5.0f < 1e-6) printf("supersonic\n");
  10. 	else printf("hypersonic\n");
  11. }
  12.  
  13. int main() {
  14. 	double v, T;
  15. 	scanf("%lf %lf", &v, &T);
  16. 	mach(v, T);
  17. 	return 0;
  18. }

【IO】气体扩散

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. double rate(double m1, double m2) {
  5. 	return sqrt(m2 / m1);
  6. }
  7.  
  8. int main() {
  9. 	double m1, m2;
  10. 	scanf("%lf %lf", &m1, &m2);
  11. 	printf("%.4lf\n", rate(m1, m2));
  12. 	return 0;
  13. }

【字符串】左右操作

  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. void quickSort(char str[], int left, int right) {
  5. 	if (left >= right) return;
  6. 	char flag = str[(left + right) / 2];
  7. 	int head = left - 1, tail = right + 1;
  8. 	while (head < tail) {
  9. 		do head++; while (str[head] > flag);
  10. 		do tail--; while (str[tail] < flag);
  11. 		if (head < tail) {
  12. 			char tmp = str[head];
  13. 			str[head] = str[tail], str[tail] = tmp;
  14. 		}
  15. 	}
  16. 	quickSort(str, left, tail);
  17. 	quickSort(str, tail + 1, right);
  18. }
  19.  
  20. void reverse(char str[], int begin, int end) {
  21. 	int head = begin, tail = end;
  22. 	while (head <= tail) {
  23. 		char tmp = str[head];
  24. 		str[head] = str[tail], str[tail] = tmp;
  25. 		++head, --tail;
  26. 	}
  27. }
  28.  
  29. int main() {
  30. 	char str[1005] = "";
  31. 	scanf("%s", str);
  32. 	int len = strlen(str);
  33. 	quickSort(str, 0, len / 2 - 1);
  34. 	int mid = (len & 1) ? (len /2 + 1) : (len / 2);
  35. 	reverse(str, mid, len - 1);
  36. 	printf("%s", str);
  37. 	return 0;
  38. }

【结构体】空中交通管制

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4.  
  5. typedef struct {
  6. 	char id[64];
  7. 	int x;
  8. 	int y;
  9. } Plane;
  10.  
  11. int main() {
  12. 	int n;
  13. 	scanf("%d", &n);
  14. 	Plane *plane[n];
  15. 	for (int i = 0; i < n; ++i) {
  16. 		plane[i] = (Plane *) malloc(sizeof(Plane));
  17. 		scanf("%s %d %d", plane[i]->id, &plane[i]->x, &plane[i]->y);
  18. 	}
  19.  
  20. 	double minDist = 1e9;
  21. 	int idx1, idx2;
  22. 	for (int i = 0; i < n - 1; ++i) {
  23. 		for (int j = i + 1; j < n; ++j) {
  24. 			double dist = sqrt(pow(plane[i]->x - plane[j]->x, 2) +
  25. 				pow(plane[i]->y - plane[j]->y, 2));
  26. 			if (dist < minDist) idx1 = i, idx2 = j, minDist = dist;
  27. 		}
  28. 	}
  29. 	printf("%s-%s %.4lf", plane[idx1]->id, plane[idx2]->id, minDist);
  30. 	return 0;
  31. }
  32.  
  33. /*
  34. 6
  35. UA057 2 3
  36. AA044 12 30
  37. BA1534 40 50
  38. DL262 5 1
  39. AF001 12 10
  40. SK837 3 4
  41. */

【数组】重复元素

  1. #include <stdio.h>
  2.  
  3. int arr[1005] = {0};
  4.  
  5. int main() {
  6. 	int n, cnt = 0;
  7. 	scanf("%d", &n);
  8. 	for (int i = 0; i < n; ++i)	{
  9. 		scanf("%d", &arr[i]);
  10. 		for (int j = 0; j < i; ++j)
  11. 			if (arr[i] == arr[j]) {
  12. 				++cnt;
  13. 				break;
  14. 			}
  15. 	}
  16. 	printf("%d\n", cnt);
  17. }
  18.  
  19. //10 1 10 20 1 25 1 10 30 25 1

【文件】平方根

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. int main() {
  5. 	int n;
  6. 	scanf("%d", &n);
  7. 	FILE *fp1 = fopen("rr.dat", "w");
  8. 	for (int i = 1; i <= n; ++i)
  9. 		fprintf(fp1, "%.6lf ", sqrt(i));
  10. 	fclose(fp1);
  11.  
  12. 	FILE *fp2 = fopen("rr.dat", "r");
  13. 	for (int i = 1; i <= n; ++i) {
  14. 		double output;
  15. 		fscanf(fp2, "%lf", &output);
  16. 		printf("%.6lf ", output);
  17. 	}
  18. 	fclose(fp2);
  19. 	return 0;
  20. }

 【算法】零钞

  1. #include <stdio.h>
  2.  
  3. int main() {
  4. 	int s;
  5. 	scanf("%d", &s);
  6.  
  7. 	int cnt[4], r = s;
  8. 	cnt[0] = r / 10, r -= cnt[0] * 10;
  9. 	cnt[1] = r / 5, r -= cnt[1] * 5;
  10. 	cnt[2] = r / 2, cnt[3] = r - cnt[2] * 2;
  11.  
  12. 	if (cnt[3]) printf("1=%d\n", cnt[3]);
  13. 	if (cnt[2]) printf("2=%d\n", cnt[2]);
  14. 	if (cnt[1]) printf("5=%d\n", cnt[1]);
  15. 	if (cnt[0]) printf("10=%d\n", cnt[0]);
  16. 	return 0;
  17. }

【枚举】机场翻牌显示

  1. #include <stdio.h>
  2. #include <ctype.h>
  3.  
  4. int cnt(char src, char dest) {
  5. 	if (src == dest) return 0;
  6. 	if (isupper(src))
  7. 		return (src < dest) ? (dest - src) : ('Z' - src + dest - 'A' + 1);
  8. 	if ('0' <= src && src <= '9') {
  9. /*		if (src == '0') src = '9' + 1;
  10. 		if (dest == '0') dest = '9' + 1;
  11. 		return (src > dest) ? (src - dest) : (src - '0' + '9' + 1 - dest + 1);*/
  12. 		return (src < dest) ? (dest - src) : ('9' - src + dest - '0' + 1);
  13. 	}
  14. 	return -1;
  15. }
  16.  
  17. int main() {
  18. 	char id1[10] = "", id2[10] = "";
  19. 	scanf("%s %s", id1, id2);
  20. 	int num = 0;
  21. 	for (int i = 0; id1[i] && id2[i] ; ++i) num += cnt(id1[i], id2[i]);
  22. 	printf("%d\n", num);
  23. 	return 0;
  24. }

【递归】阿克曼数

  1. #include <stdio.h>
  2.  
  3. int ack(int m, int n) {
  4. 	if (m == 0) return n + 1;
  5. 	if (n == 0) return ack(m - 1, 1);
  6. 	return ack(m - 1, ack(m, n -1));
  7. }
  8.  
  9. int main() {
  10. 	int m, n;
  11. 	scanf("%d %d", &m, &n);
  12. 	printf("%d\n", ack(m, n));
  13. 	return 0;
  14. }

声明：本文内容由网友自发贡献，不代表【wpsshop博客】立场，版权归原作者所有，本站不承担相应法律责任。如您发现有侵权的内容，请联系我们。转载请注明出处：https://www.wpsshop.cn/w/羊村懒王/article/detail/271774
推荐阅读
相关标签

Copyright © 2003-2013 www.wpsshop.cn 版权所有，并保留所有权利。

  

闽ICP备14008679号