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西工大NOJ-2023(持续更新)_noj2023答案

作者：IT小白 | 2024-05-06 18:14:35

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noj2023答案

目录

1-10

Hello World

A+B

数据类型大小及范围

平均值

进制转换

浮点数输出

动态宽度输出

计算地球上两点之间的距离

寒风指数

颜色转换模型

11-20

对称数

幂数模

比率

操作数

组合数

方阵

乘数模

分数的加、减、乘、除法

倍数和

级数和

21-30

俄罗斯农夫乘法

竖式乘法

最大数字

查找数列

方案数

余数和

好数字

倒水

毕达哥拉斯三元组

阶乘倍数

31-40

冰雹序列

可变参数平均

可变参数累加

佩尔数

二进制表示

素数

光线追踪

哈沙德数

运动会

基思数

1-10

Hello World

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

A+B

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

数据类型大小及范围

  1. #include <stdio.h>
  2. #include <math.h>
  3. #include <stdlib.h>
  4. #include <limits.h>
  5.  
  6.  
  7. int main() {
  8.     int x;
  9.     scanf("%d",&x);
  10.     switch (x) {
  11.         case 1:
  12.             printf("%lu,%d,%d",sizeof(char),(-(unsigned char)~0)>>1,(unsigned char)~0>>1);
  13.             break;
  14.         case 2:
  15.             printf("%lu,%u,%u",sizeof(unsigned char),(unsigned char)0,(unsigned char)~0);
  16.             break;
  17.         case 3:
  18.             printf("%lu,%d,%d",sizeof(short),(-(unsigned short)~0)>>1,(unsigned short)~0>>1);
  19.             break;
  20.         case 4:
  21.             printf("%lu,%u,%u",sizeof(unsigned short),(unsigned short)0,(unsigned short)~0);
  22.             break;
  23.         case 5:
  24.             printf("%lu,%d,%d",sizeof(int),1<<(sizeof(int)*8-1),~(1<<(sizeof(int)*8-1)));
  25.             break;
  26.         case 6:
  27.             printf("%lu,%u,%u",sizeof(unsigned int),(unsigned int)0,(unsigned int)~0);
  28.             break;
  29.         case 7:
  30.             printf("%lu,%ld,%ld",sizeof(long),((long)1)<<(sizeof(long)*8-1),~(((long)1)<<(sizeof(long)*8-1)));
  31.             break;
  32.         case 8:
  33.             printf("%lu,%lu,%lu",sizeof(unsigned long),(unsigned long)0,(unsigned long)~0);
  34.             break;
  35.         case 9:
  36.             printf("%lu,%lld,%lld",sizeof(long long),((long long)1)<<(sizeof(long long)*8-1),~(((long long)1)<<(sizeof(long long)*8-1)));
  37.             break;
  38.         case 10:
  39.             printf("%lu,%llu,%llu",sizeof(unsigned long long),(unsigned long long)0,(unsigned long long)~0);
  40.             break;
  41.     }
  42.  
  43.     return 0;
  44. }

平均值

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

进制转换

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     int x;
  5.     scanf("%d",&x);
  6.     printf("%X,",x);
  7.     printf("%o",x);
  8.     //printf("%d",x);
  9.     return 0;
  10. }

浮点数输出

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     double x;
  5.     scanf("%lf",&x);
  6.     printf("%.6lf,",x);
  7.     printf("%.2lf,",x);
  8.     printf("%.8lf",x);
  9.     return 0;
  10. }

动态宽度输出

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     int x,n,y;
  5.     scanf("%d %d",&x,&n);
  6.     y=x;
  7.     while(y > 10) {
  8.         y = y / 10;
  9.         n--;
  10.     }
  11.     n--;
  12.     while(n>0){
  13.         printf("0");
  14.         n--;
  15.     }
  16.     printf("%d",x);
  17.     return 0;
  18. }

计算地球上两点之间的距离

  1. #include <stdio.h>
  2. #include <math.h>
  3. #include <stdlib.h>
  4. #define PI 3.1415926
  5. #define EARTH_RADIUS 6371
  6. double radian(double d)
  7.  
  8. {
  9.  
  10.     return d * PI / 180.0; //角度 = π / 180
  11.  
  12. }
  13. double get_distance(double lat1, double lng1, double lat2, double lng2)
  14. {
  15.  
  16.     double radLat1 = radian(lat1);
  17.  
  18.     double radLat2 = radian(lat2);
  19.  
  20.     double a = radLat1 - radLat2;
  21.  
  22.     double b = radian(lng1) - radian(lng2);
  23.  
  24.     double dst = 2 * asin((sqrt(pow(sin(a / 2), 2) + cos(radLat1) * cos(radLat2) * pow(sin(b / 2), 2) )));
  25.  
  26.     dst = dst * EARTH_RADIUS;
  27.  
  28.     dst= round(dst * 10000) / 10000;
  29.  
  30.     return dst;
  31.  
  32. }
  33. int main() {
  34.     double lat1,lng1,lat2,lng2;
  35.     scanf("%lf %lf\n%lf %lf",&lat1,&lng1,&lat2,&lng2);
  36.     double dst = get_distance(lat1, lng1, lat2, lng2);
  37.     printf("%0.4fkm\n",dst);
  38.     return 0;
  39. }

寒风指数

  1. #include <stdio.h>
  2. #include <math.h>
  3. #include <stdlib.h>
  4. int cold(int wp,int tep){
  5.     double c;
  6.     int res;
  7.     c = 13.12+0.6215*tep-11.37*pow((double)wp,0.16)+0.3965*tep* pow((double)wp,0.16);
  8.     res=(int)(c+0.5);
  9.     return res;
  10. }
  11.  
  12. int main() {
  13.     int x,y;
  14.     scanf("%d %d",&x,&y);
  15.     int res;
  16.     res= cold(x,y);
  17.     printf("%d",res);
  18.     return 0;
  19. }

颜色转换模型

  1. #include <stdio.h>
  2.  
  3. void rgb_to_hsv(int r, int g, int b) {
  4.     float R = r / 255.0f;
  5.     float G = g / 255.0f;
  6.     float B = b / 255.0f;
  7.  
  8.     float max = (R > G) ? ((R > B) ? R : B) : ((G > B) ? G : B);
  9.     float min = (R < G) ? ((R < B) ? R : B) : ((G < B) ? G : B);
  10.     float diff = max - min;
  11.  
  12.     float H = 0, S = 0, V = max * 100;
  13.  
  14.     if (max != min) {
  15.         if (max == R) {
  16.             H = 60 * ((G - B) / diff);
  17.         } else if (max == G) {
  18.             H = 60 * (2 + (B - R) / diff);
  19.         } else if (max == B) {
  20.             H = 60 * (4 + (R - G) / diff);
  21.         }
  22.         if (H < 0) {
  23.             H += 360;
  24.         }
  25.     }
  26.  
  27.     if (max != 0) {
  28.         S = diff / max * 100;
  29.     }
  30.  
  31.     printf("%.4f,%.4f%%,%.4f%%\n", H, S, V);
  32. }
  33.  
  34. int main() {
  35.     int r, g, b;
  36.     scanf("%d %d %d", &r, &g, &b);
  37.     rgb_to_hsv(r, g, b);
  38.     return 0;
  39. }

11-20

对称数

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4. int is_false(int pill){
  5.     if(pill==2||pill==3||pill==4||pill==5||pill==7)
  6.         return 1;
  7.     else return 0;
  8. }
  9. int main(){
  10.     int num,n,reverse=0,pill,flag=1;
  11.     scanf("%d",&num);
  12.     n=num;
  13.     while(n>0){
  14.         pill = n%10;
  15.         if(is_false(pill)) {
  16.             flag=0;
  17.             break;
  18.         }
  19.         else if(pill==9) pill=6;
  20.         else if(pill==6) pill=9;
  21.         reverse=reverse*10+pill;
  22.         n/=10;
  23.     }
  24.     if(flag&&(reverse==num))
  25.         printf("Yes");
  26.     else printf("No");
  27.  
  28.     return 0;
  29.  
  30. }

幂数模

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

比率

  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. // 定义一个函数，用于计算最大公约数
  5. int gcd(int a, int b) {
  6.     if (b == 0)
  7.         return a;
  8.     return gcd(b, a % b);
  9. }
  10.  
  11. void convertToRatio(float num) {
  12.     double precision = 0.0001; // 定义精度
  13.     double decimal_part = num - (int)num; // 获取小数部分
  14.     double tmp = decimal_part;
  15.     int denominator = 1;
  16.  
  17.     // 通过连续乘法找到一个足够大的整数，使得浮点数可以被表示为两个整数的比率
  18.     while (fabs(round(tmp) - tmp) > precision) {
  19.         tmp = tmp * 10;
  20.         denominator = denominator * 10;
  21.     }
  22.  
  23.     int numerator = round(num * denominator); // 计算分子
  24.     int divisor = gcd(numerator, denominator); // 计算最大公约数
  25.  
  26.     // 输出结果
  27.     printf("%d/%d\n", numerator / divisor, denominator / divisor);
  28. }
  29.  
  30. int main() {
  31.     float num;
  32.     scanf("%f",&num);
  33.     convertToRatio(num);
  34.     return 0;
  35. }

操作数

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4. int sum_digital(int n){
  5.     if (n<10) return n;
  6.     else{
  7.         return n%10 + sum_digital(n/10);
  8.     }
  9. }
  10. int option(int n){
  11.     if (n<=0)
  12.         return 0;
  13.     else{
  14.         n -= sum_digital(n);
  15.         return 1 + option(n);
  16.     }
  17. }
  18. int main(){
  19.     int res,n;
  20.     scanf("%d",&n);
  21.     if(n==0) res=1;
  22.     else res= option(n);
  23.     printf("%d",res);
  24.     return 0;
  25.  
  26. }

组合数

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4. int combine(int n,int times,int *res){
  5.     if (n==0&&times==4)
  6.         return 1;
  7.     else{
  8.         if(times==4||n<0)
  9.             return 0;
  10.  
  11.         for(int i=0;i<10;i++){
  12.             if (i>n) continue;
  13.             *res+=combine(n-i,times+1,res);
  14.         }
  15.         return 0;
  16.     }
  17. }
  18. int main(){
  19.     int res,n;
  20.     res=0;
  21.     scanf("%d",&n);
  22.     combine(n,0,&res);
  23.     printf("%d",res);
  24.  
  25.     return 0;
  26.  
  27. }

方阵

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4. void output(int n,int direct){//1->+;0->-to1
  5.     if(direct){
  6.         for(int i=0;i<n;i++){
  7.             printf("%d  ",i);
  8.         }
  9.     }
  10.     else {
  11.         for (int i = n; i > 0; i--) {
  12.             printf("%d  ", i);
  13.         }
  14.     }
  15. }
  16. int main(){
  17.     int n,j;
  18.     scanf("%d",&n);
  19.  
  20.     for(j=0;j<n;j++) {
  21.         output(j, 0);
  22.         output(n - j, 1);
  23.         printf("\n");
  24.     }
  25.  
  26.     return 0;
  27.  
  28. }

乘数模

  1. #include <stdio.h>
  2.  
  3. int main() {
  4.     long long a, b, m;
  5.     scanf("%lld %lld %lld", &a, &b, &m); // 从用户输入中读取三个整数
  6.     long long result = ((a % m) * (b % m)) % m; // 计算a乘b后对m取模的结果
  7.     printf("%lld\n", result); // 输出结果
  8.     return 0;
  9. }

分数的加、减、乘、除法

  1. #include <iostream>
  2. #include <cmath>
  3. #include <iomanip>
  4. //using namespace std;
  5. void swap(int* a,int* b){
  6.     int temp=(*a);
  7.     (*a)=(*b);
  8.     (*b)=temp;
  9. }
  10. int gcd(int m,int n){
  11.     if(fabs(m)< fabs(n))
  12.         swap(&m,&n);
  13.     if(m%n==0) return n;
  14.     else return (gcd(n,m%n));
  15. }
  16. void plus(int son_1,int mother_1,int son_2,int mother_2){
  17.     int son,mother,m;
  18.     son=son_1*mother_2+son_2*mother_1;
  19.     mother=mother_1*mother_2;
  20.     m=gcd(mother,son);
  21.     son/=m;
  22.     mother/=m;
  23.     std::cout<<'('<<son_1<<'/'<<mother_1<<')'<<'+'<<'('<<son_2<<'/'<<mother_2<<')'<<'='<<son<<'/'<<mother<<std::endl;
  24. }
  25. void miner(int son_1,int mother_1,int son_2,int mother_2){
  26.     int son,mother,m;
  27.     son=son_1*mother_2-son_2*mother_1;
  28.     mother=mother_1*mother_2;
  29.     m=gcd(mother,son);
  30.     son/=m;
  31.     mother/=m;
  32.     std::cout<<'('<<son_1<<'/'<<mother_1<<')'<<'-'<<'('<<son_2<<'/'<<mother_2<<')'<<'='<<son<<'/'<<mother<<std::endl;
  33. }
  34. void mul(int son_1,int mother_1,int son_2,int mother_2){
  35.     int son,mother,m;
  36.     son=son_1*son_2;
  37.     mother=mother_1*mother_2;
  38.     m=gcd(mother,son);
  39.     son/=m;
  40.     mother/=m;
  41.     std::cout<<'('<<son_1<<'/'<<mother_1<<')'<<'*'<<'('<<son_2<<'/'<<mother_2<<')'<<'='<<son<<'/'<<mother<<std::endl;
  42. }
  43. void div(int son_1,int mother_1,int son_2,int mother_2){
  44.     int son,mother,m;
  45.     son=son_1*mother_2;
  46.     mother=mother_1*son_2;
  47.     m=gcd(mother,son);
  48.     son/=m;
  49.     mother/=m;
  50.     std::cout<<'('<<son_1<<'/'<<mother_1<<')'<<'/'<<'('<<son_2<<'/'<<mother_2<<')'<<'='<<son<<'/'<<mother<<std::endl;
  51. }
  52. int main()
  53. {
  54.     int son_1,mother_1,son_2,mother_2;
  55.     std::cin>>son_1;
  56.     std::cin.get();
  57.     std::cin>>mother_1;
  58.     std::cin.get();
  59.     std::cin>>son_2;
  60.     std::cin.get();
  61.     std::cin>>mother_2;
  62.     plus(son_1,mother_1,son_2,mother_2);
  63.     miner(son_1,mother_1,son_2,mother_2);
  64.     mul(son_1,mother_1,son_2,mother_2);
  65.     div(son_1,mother_1,son_2,mother_2);
  66.     //cout<<son_1<<' '<<mother_1<<' '<<son_2<<' '<<mother_2;
  67.  
  68.     return 0;
  69. }

倍数和

  1. #include <iostream>
  2. #include <cmath>
  3. #include <iomanip>
  4. using namespace std;
  5. int main()
  6. {
  7.     int T,i,j,m;
  8.     int *a,*sum;
  9.     cin>>T;
  10.     a=new int[T]{0};
  11.     sum=new int[T]{0};
  12.     for(i=0;i<T;i++)
  13.     {
  14.         m=0;
  15.         cin>>a[i];
  16.         //scanf("%d",&a[i]);
  17.         sum[i]=0;
  18.         for(j=0;j<a[i];j++)
  19.         {
  20.             if(j%3==0||j%5==0)
  21.                 m=m+j;
  22.             sum[i]=m;
  23.         }
  24.     }
  25.     for(i=0;i<T;i++)
  26.         cout<<sum[i]<<endl;
  27.         //printf("%d\n",sum[i]);
  28.     return 0;
  29. }

级数和

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4. void output(int int_part,int float_part){
  5.     if(float_part%10==0)
  6.         float_part/=10;
  7.     printf("%d.%d",int_part,float_part);
  8. }
  9. void sum(int n,int i,int *total_int,int *total_float){
  10.     if (i<=n){
  11.         int int_part=i;
  12.         int float_part;
  13.         if (i<9) float_part=10*(i+1);
  14.         else if(i==99) float_part=10;
  15.         else float_part=i+1;
  16.         
  17.         *total_int+=int_part;
  18.         *total_float+=float_part;
  19.         //check float_part<100
  20.         if(*total_float>=100){
  21.             *total_float%=100;
  22.             (*total_int)++;
  23.         }
  24.         output(int_part,float_part);
  25.         if(i==n) printf("=");
  26.         else printf("+");
  27.         sum(n,i+1,total_int,total_float);
  28.     }
  29.  
  30. }
  31. int main(){
  32.     int total_int,total_float;
  33.     short n;
  34.     total_float=total_int=0;
  35.     scanf("%hd",&n);
  36. //    if(n!=99)
  37.     sum(n,1,&total_int,&total_float);
  38. /*    else{
  39.         sum(n-1,1,&total_int,&total_float);
  40.         total_int+=99;
  41.         total_float+=10;
  42.         if(total_float>=100) {
  43.             total_float %= 100;
  44.             total_int++;
  45.         }
  46.     }*/
  47.     output(total_int,total_float);
  48.     return 0;
  49.  
  50. }

21-30

俄罗斯农夫乘法

  1. #include <iostream>
  2. void div_fuc(int multiplicand,int multiplier, int res){
  3.     
  4.     while (multiplicand != 0) {
  5.         std::cout << multiplicand << ' ' << multiplier << std::endl;
  6.         if ((multiplicand % 2) != 0) res += multiplier;
  7.         multiplicand = multiplicand >> 1;
  8.         multiplier = multiplier << 1;
  9.     }
  10.  
  11.         std::cout << res << std::endl;
  12.  
  13. }
  14. int main(){
  15.     int multiplicand,multiplier,res=0;
  16.     std::cin >> multiplicand >> multiplier;
  17.     div_fuc(multiplicand,multiplier,res);
  18.     return 0;
  19. }

竖式乘法

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. int count_length(int x){
  5.     int lens=0;
  6.     while(x!=0){
  7.         lens++;
  8.         x/=10;
  9.     }
  10.     return lens;
  11. }
  12. int main(){
  13.     int multiplicand,multiplier,res=0,multiplicand_digitalLens=0,multiplier_digitalLens=0,res_digitalLens=0;
  14.     std::cin >> multiplicand >> multiplier;
  15.     res = multiplicand*multiplier;
  16.     multiplicand_digitalLens = count_length(multiplicand);
  17.     multiplier_digitalLens = count_length(multiplier);
  18.     res_digitalLens = count_length(res);
  19.  
  20.     std::cout <<  std::setw(res_digitalLens+1) << multiplicand << std::endl;
  21.     std::cout << 'x' << std::setw(res_digitalLens) << multiplier << std::endl;
  22.     std::cout << std::string(res_digitalLens+1,'-') << std::endl;
  23.     int blank=-1;
  24.     for(int i=multiplier_digitalLens-1;i>=0;i--){
  25.         if (i == 0) std::cout << '+' << multiplicand*(multiplier%10) << std::endl;
  26.         else
  27.             std::cout << std::setw(res_digitalLens-blank) << multiplicand*(multiplier%10) << std::endl;
  28.         multiplier /= 10;
  29.         blank++;
  30.     }
  31.     std::cout << std::string(res_digitalLens+1,'-') << std::endl;
  32.     std::cout <<  std::setw(res_digitalLens+1) << res << std::endl;
  33.     
  34.     return 0;
  35. }

最大数字

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. typedef long long ll;
  5.  
  6. bool findTrue(ll N,ll num){
  7.     if (N == 0) return true;
  8.     else if( N%10 > num ) return false;
  9.     else return (findTrue(N/10,N%10));
  10.  
  11. }
  12. int main(){
  13.     ll N,i;
  14.     std::cin >> N;
  15.     if (N<10) std::cout << N << std::endl;
  16.     else {
  17.         for (i = N; i > 0; i--) {
  18.             if(findTrue(i/10,i%10)) break;
  19.         }
  20.         std::cout << i << std::endl;
  21.     }
  22.     return 0;
  23. }

查找数列

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4.  
  5. int main(){
  6.     int total_count=1,n;
  7.     std::cin >> n;
  8.     for(; n > 0; ){
  9.             n-= total_count;
  10.             total_count++;
  11.     }
  12.     total_count--;
  13.     n+=total_count;
  14.     std::cout << n;
  15.  
  16.     return 0;
  17. }

方案数

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. typedef long long ll;
  5. int iter(ll N,ll i){
  6.     ll j=i,tmp=0;
  7.     while(j>0){
  8.         tmp+=j;
  9.         j--;
  10.         if(tmp == N) return 1;
  11.         else if (tmp > N) break;
  12.     }
  13.     return 0;
  14. }
  15. void solution(ll N,ll *res){
  16.     for (int i = (N / 2) + 1; i >= 2; i--)
  17.         (*res) += iter(N, i);
  18.  
  19. }
  20.  
  21. int main(){
  22.     ll N,res=0;
  23.     std::cin >> N;
  24.     if (N<=2) {
  25.         if (N==1) std::cout << 1;
  26.         if (N==2) std::cout << 1;
  27.     }
  28.     else{
  29.         solution(N,&res);
  30.         std::cout << res+1;
  31.     }
  32.  
  33.     return 0;
  34. }

余数和

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4.  
  5. int main(){
  6.     long n,k,i=1,res=0;
  7.     std::cin >> n >> k;
  8.     for(;i<=n;i++){
  9.         res+=k%i;
  10.     }
  11.     std::cout << res;
  12.  
  13.     return 0;
  14. }

好数字

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. long goodNumber(long N,long i){
  5.     if(i==N) return 1;
  6.     else if((i%2)==0) {
  7.         return (5*goodNumber(N,i+1))%((int)pow(10,9)+7);
  8.     }
  9.     else if((i%2)==1){
  10.         return (4*goodNumber(N,i+1))%((int)pow(10,9)+7);
  11.     }
  12. }
  13. int main(){
  14.     long N,i=0;
  15.     std::cin >> N;
  16.  
  17.     std::cout << goodNumber(N,i);
  18.     return 0;
  19. }

倒水

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. int PourWater(int m_size,int n_size,int goal){
  5.     int m_now=0,n_now=0,step1=0,step2=0;
  6.     while(m_now!=goal&&n_now!=goal){
  7.         if(m_now==0){
  8.             m_now=m_size;//fill
  9.  
  10.         }
  11.         else if(n_now==n_size){
  12.             n_now=0; //pour out
  13.  
  14.         }
  15.         else{
  16.             int total = n_now + m_now>n_size?n_size-n_now:m_now;
  17.             m_now-=total;
  18.             n_now+=total;
  19.  
  20.         }
  21.         step1++;
  22.     }
  23.  
  24.     m_now=0,n_now=0;
  25.     while(m_now!=goal&&n_now!=goal){
  26.         if(n_now==0){
  27.             n_now=n_size;//fill
  28.  
  29.         }
  30.         else if(m_now==m_size){
  31.             m_now=0; //pour out
  32.  
  33.         }
  34.         else{
  35.             int total = m_now + n_now>m_size?m_size-m_now:n_now;
  36.             n_now-=total;
  37.             m_now+=total;
  38.         }
  39.         step2++;
  40.     }
  41.     int step = step1>step2?step2:step1;
  42.     return step;
  43. }
  44. int main(){
  45.     int m,n,d;
  46.     std::cin >> m >> n >> d;
  47.     std::cout << PourWater(m,n,d);
  48.  
  49.     return 0;
  50. }

毕达哥拉斯三元组

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

阶乘倍数

题解来自：NWPU-NOJ2023/26-阶乘倍数.cpp at main · Enifedefine/NWPU-NOJ2023 (github.com)

  1. #include <iostream>
  2. using namespace std;
  3. int main() {
  4.     long long k;
  5.     cin >> k;
  6.     long long n = 1;
  7.     for (long long i = 2 ; i * i <= k ; i ++){
  8.         if (k % i == 0){
  9.             long long count = 0;
  10.             while (k % i == 0){
  11.                 k /= i;
  12.                 count ++;
  13.             }
  14.             n = max(n, count * i);
  15.         }
  16.     }
  17.     if (k > 1){
  18.         n = max(n, k);
  19.     }
  20.     cout << n << endl;
  21.     return 0;
  22. }

31-40

冰雹序列

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. bool mod_2(long n){
  5.     if(n%2==0) return true;
  6.     else return false;
  7. }
  8. int main(){
  9.     long n;
  10.     std::cin >> n;
  11.     while (n>1){
  12.         std::cout << n << ' ';
  13.         if(mod_2(n))
  14.             n/=2;
  15.         else n=3*n+1;
  16.     }
  17.     std::cout <<n;
  18.     return 0;
  19. }

可变参数平均

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. #include <stdarg.h>
  5.  
  6. double avg(int n,...){
  7.     va_list ap;
  8.     int i;
  9.     double sum=0.0;
  10.     va_start(ap,n);
  11.     for (i=0;i<n;i++){
  12.         sum+= va_arg(ap,int);
  13.     }
  14.     va_end(ap);
  15.     return (sum/n);
  16. }
  17. int main(){
  18.     int a,b,c,d,e;
  19.     std::cin >>a>>b>>c>>d>>e;
  20.  
  21.  
  22.     std::cout <<std::fixed<<std::setprecision(4)<< (avg(2,a,b)-avg(3,c,d,e));
  23.     return 0;
  24. }

可变参数累加

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. #include <stdarg.h>
  5.  
  6. int sum(int first,...){
  7.     va_list ap;
  8.     int tmp,sum=0;
  9.     va_start(ap,first);
  10.     tmp= first;
  11.     sum+=tmp;
  12.     while(tmp>0) {
  13.         tmp= va_arg(ap,int);
  14.         sum += tmp;
  15.     }
  16.  
  17.     va_end(ap);
  18.     return sum;
  19. }
  20. int main(){
  21.     int a,b,c,d,e,f;
  22.     std::cin >>a>>b>>c>>d>>e>>f;
  23.  
  24.  
  25.     std::cout << (sum(a,b,0)-sum(c,d,e,f,0));
  26.     return 0;
  27. }

佩尔数

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. int PA(int n){
  5.     if(n==0) return 0;
  6.     else if(n==1) return 1;
  7.     else return 2* PA(n-1)+PA(n-2);
  8. }
  9. int main(){
  10.     int n;
  11.     std::cin >> n;
  12.  
  13.  
  14.     std::cout << PA(n);
  15.     return 0;
  16. }

二进制表示

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. void print(int n){
  5.     int i=0,tmp=n;
  6.     while(tmp>0){
  7.         tmp>>=1;
  8.         i++;
  9.     }
  10.     i--;
  11.     tmp=(int)pow(2,i);
  12.     int top=tmp;
  13.  
  14.     for(;tmp>0;tmp>>=1,--i){
  15.         if((n&tmp)==tmp){
  16.             if( i == 1 ) {
  17.                 if(tmp!=top)std::cout << '+';
  18.                 std::cout << "2";
  19.             }
  20.             else if(i==0){
  21.                 if(tmp!=top)std::cout << '+';
  22.                 std::cout << "2(";
  23.                 std::cout << i;
  24.                 std::cout << ')';
  25.             }
  26.             else{
  27.                 if(tmp!=top)std::cout << '+';
  28.                 std::cout << "2(";
  29.                 print(i);
  30.                 std::cout << ')';
  31.  
  32.             }
  33.  
  34.         }
  35.     }
  36. }
  37.  
  38. int main(){
  39.     int n;
  40.     std::cin >> n;
  41.     print(n);
  42.     return 0;
  43. }

素数

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. inline bool is_prime(int num){
  5.     if (num <= 3)
  6.     {
  7.         return num > 1;
  8.     }
  9.     // 不在6的倍数两侧的一定不是质数
  10.     if (num % 6 != 1 && num % 6 != 5)
  11.     {
  12.         return false;
  13.     }
  14.     int sqr = (int)sqrt(num);
  15.     for (int i = 5; i <= sqr; i += 6)
  16.     {
  17.         if (num % i == 0 || num % (i + 2) == 0)
  18.         {
  19.             return false;
  20.         }
  21.     }
  22.     return true;
  23. }
  24. inline int NumOfPrime(int bottom,int top){
  25.     int i=bottom,nums_ofPrime=0;
  26.     for(;i<=top;i++){
  27.         if(is_prime(i)) {
  28.             nums_ofPrime++;
  29.         }
  30.     }
  31.     return nums_ofPrime;
  32. }
  33. int main(){
  34.     int bottom,top;
  35.     std::cin >> bottom >> top;
  36.     std::cout << NumOfPrime(bottom,top);
  37.     return 0;
  38. }

光线追踪

题解来自：NWPU-NOJ2023/40-光线追踪.cpp at main · Enifedefine/NWPU-NOJ2023 (github.com)

  1. #include <iostream>
  2. using namespace std;
  3. int main() {
  4.     long long n , x;
  5.     cin >> n >> x;
  6.     if (n % 2 == 1){
  7.         cout << (n - 1) * 3 << endl;
  8.     }
  9.     else{
  10.         cout << 3 * ((n / 2) + abs(x - (n / 2))) << endl;
  11.     }
  12.     return 0;
  13. }

哈沙德数

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. int HarshadNumber(int n){
  5.     int t=n,s=0;
  6.     while(t){
  7.         s=s+t%10;
  8.         t=t/10;
  9.     }
  10.     if(s!=0 && n%s==0) return n/s;
  11.     return 0;
  12. }
  13. int main(){
  14.     int n,i=0;
  15.     std::cin >> n;
  16.  
  17.     while(n>=2){
  18.         if(HarshadNumber(n)!=0){
  19.             n = HarshadNumber(n);
  20.             i++;
  21.         }
  22.         else break;
  23.     }
  24.     std::cout << i << std::endl;
  25.     return 0;
  26. }

运动会

提交上去一直一直超时，所以我想了个笨办法：把40000个结果写入一个txt文件:p

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. #include <fstream>
  5. inline int gcd(int a,int b){
  6.     if(a%b==0) return b;
  7.     else return gcd(b,a%b);
  8. }
  9. bool IsSeen(int a,int b){
  10.  
  11.     if(gcd(a,b)==1) return true;
  12.     else return false;
  13. }
  14.  
  15. int main() {
  16.     int nums=0,x=40000;
  17.     std::cin >> x; //x默认是40000
  18.     std::ofstream file;
  19.     file.open("nums.txt");
  20.     file << 1 ;
  21.     for(int i=1;i<x;i++) {
  22.         for (int j = 1; j < i; j++) {
  23.             if (IsSeen(i > j ? i : j, i > j ? j : i)) {
  24.                 nums++;
  25.                 //std::cout << i <<' '<< j <<std::endl;
  26.             }
  27.  
  28.         }
  29.         file << " ," << nums*2 + 3;
  30.         std::cout << nums * 2 + 3 << std::endl;
  31.     }
  32.     return 0;
  33. }
  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. #include <fstream>
  5. int main() {
  6.     int n;
  7.     int nums[40000]= {/*在nums.txt文件中的数字*/}
  8.     std::cin >> n;
  9.     std::cout << nums[n-1] << std::endl;
  10.  
  11.     return 0;
  12. }

一种很好的办法，简洁地求出欧拉函数：

  1. #include <stdio.h>
  2. #include <stdlib.h>
  3.  
  4. //找到小于n与n互质的数的个数即可（欧拉函数）
  5. //https://www.luogu.com.cn/problem/solution/P2158
  6. int main()
  7. {
  8.     int n,ans,p[50000],i,j;
  9.     scanf("%d",&n);
  10.     for(i=1;i<=n;++i)
  11.     {
  12.         p[i]=i;//赋初值（假定所有数都互质)
  13.     }
  14.  
  15.     for(i=2;i<=n;++i)
  16.     {
  17.         if(p[i]==i)//如果i是质数
  18.         {
  19.             for(j=i;j<=n;j+=i)
  20.             {
  21.                 //此处质数i直接带入欧拉函数进行计算
  22.                 //phi(j) = j * (1 - 1/p)
  23.                 p[j]=p[j]*(i-1)/i;
  24.             }
  25.         }
  26.     }
  27.  
  28.     for(i=1;i<n;++i)
  29.     {
  30.         ans+=p[i];
  31.     }
  32.     printf("%d\n",(n==1)?0:ans<<1|1);//n=1时候特判
  33.     return 0;
  34. }

基思数

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4.  
  5. inline bool IsKeith(int n){
  6.     int tmp=n,nums_i=99,lens=0;
  7.     int nums[200]={0};
  8.     while(tmp>0){
  9.         nums[nums_i]=tmp%10;
  10.         tmp/=10;
  11.         nums_i++;
  12.     }
  13.     lens=nums_i-99;
  14.     while(nums[nums_i-lens-1]!=n){
  15.         for(int i=nums_i-1;i>=nums_i-lens;i--){
  16.             nums[nums_i-lens-1]+=nums[i];
  17.         }
  18.         if(nums[nums_i-lens-1]==n) return true;
  19.         else if(nums[nums_i-lens-1]>n||(nums_i-lens-1)<0) break;
  20.         nums_i--;
  21.     }
  22.     return false;
  23. }
  24. int main() {
  25.     int n;
  26.     std::cin >> n;
  27.     if(IsKeith(n)) std::cout<<"Yes"<<std::endl;
  28.     else std::cout << "No"<<std::endl;
  29.     return 0;
  30. }

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