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c++ noj记录[已更新至40题]_noj习题风寒指数

作者：2023面试高手 | 2024-05-06 18:18:09

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noj习题风寒指数

用于记录完成noj时遇到的问题和对自己来说需要注意的事项，可供参考，不建议抄袭。 

1~10

 Hello World

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	cout << "Hello World";
  7. }

A+B

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int a, b;
  7. 	cin >> a >> b;
  8. 	cout << a + b;
  9. }

数据类型大小及范围

  1. ​
  2. ​
  3. #include<iostream>
  4. #include<stdio.h>
  5. #include<limits.h>
  6. using namespace std;
  7.  
  8. int main()
  9. {
  10. 	int d;
  11. 	cin >> d;
  12.  
  13. 	switch (d)
  14. 	{
  15. 	case 1:
  16. 		cout << sizeof(char) << ',' << CHAR_MIN << ',' << CHAR_MAX;
  17. 		break;
  18. 	case 2:
  19. 		cout << sizeof( unsigned char) << ',' << 0 << ',' << UCHAR_MAX;
  20. 		break;
  21. 	case 3:
  22. 		cout << sizeof(short) << ',' << SHRT_MIN << ',' << SHRT_MAX;
  23. 		break;
  24. 	case 4:
  25. 		cout << sizeof(unsigned short) << ',' << 0 << ',' << USHRT_MAX;
  26. 		break;
  27. 	case 5:
  28. 		cout << sizeof(int) << ',' << INT_MIN << ',' << INT_MAX;
  29. 		break;
  30. 	case 6:
  31. 		cout << sizeof(unsigned int) << ',' << 0 << ',' << UINT_MAX;
  32. 		break;
  33. 	case 7:
  34. 		cout << sizeof(long) << ',' << LONG_MIN << ',' << LONG_MAX;
  35. 		break;
  36. 	case 8:
  37. 		cout << sizeof(unsigned long) << ',' << 0 << ',' << ULONG_MAX;
  38. 		break;
  39. 	case 9:
  40. 		cout << sizeof(long long) << ',' << LLONG_MIN << ',' << LLONG_MAX;
  41. 		break;
  42. 	case 10:
  43. 		cout << sizeof(unsigned long long) << ',' << 0 << ',' << ULLONG_MAX;
  44. 		break;
  45.     default：；
  46. 	}
  47.  
  48. }
  49.  
  50.  
  51. ​

 使用XXX_MAX/MIN要记得用<limits.h>库

平均值

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int a, b;
  7. 	cin >> a >> b;
  8. 	cout << a + ((b - a) >>1);
  9.  
  10. }

注意不能直接使用(a+b)/2，会造成数据溢出

进制转换

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int a;
  7. 	cin >> a;
  8.  
  9. 	cout << uppercase << hex << a << "," << oct << a;
  10.  
  11. }

浮点数输出

  1. #include<iostream>
  2. #include<stdio.h>
  3. using namespace std;
  4.  
  5. int main()
  6. {
  7. 	double a;
  8. 	cin >> a;
  9. 	printf("%.6f,%.2f,%.8f", a, a, a);
  10. }

动态宽度输出

  1. #include<iostream>
  2. #include<iomanip>
  3. using namespace std;
  4.  
  5. int main()
  6. {
  7. 	int a, b;
  8. 	cin >> a >> b;
  9. 	cout << setw(b)<<setfill('0') << a;
  10. }

引用<iomanip>库后，可以使用setw(int n)设置域宽为n，使用setfill(char c)填充字符

计算地球上两点之间的距离

  1. #include<iostream>
  2. #include<math.h>
  3. using namespace std;
  4.  
  5. double lon1, lat1, lon2, lat2;
  6. const double r = 6371.0, pi = 3.1415926535;
  7.  
  8. double hav(double x)
  9. { 
  10. 	return (1 - cos(x)) / 2;
  11. }
  12.  
  13. double rad(double *x)
  14. {
  15. 	return *x = *x * pi / 180;
  16. }
  17.  
  18. int main()
  19. {
  20. 	double h,d;
  21. 	cin >> lat1 >> lon1 >>lat2 >> lon2;
  22. 	
  23. 	rad(&lat1); rad(&lon1); rad(&lat2); rad(&lon2);
  24. 	h = hav(lat2 - lat1) + cos(lat1) * cos(lat2) * hav(lon2 - lon1);
  25. 	d = acos(1 - (2 * h)) * r;
  26. 	
  27. 	printf("%.4fkm", d);
  28.  
  29. 	return 0;
  30. }
  31.

注意看题别把经纬度搞错了，变量和意义要对应上。

风寒指数

  1. #include<iostream>
  2. #include<cmath>
  3. using namespace std;
  4.  
  5. int main()
  6. {
  7. 	double v, t, c,v1;
  8.  
  9. 	cin >> v >> t;
  10. 	v1 = pow(v, 0.16);
  11. 	c = 13.12 + 0.6215 * t - (11.37 * v1) + (0.3965 * t * v1);
  12. 	printf("%.0f", c);
  13. }

 引用<cmath>后可以使用pow(x,y)函数进行指数运算，结果为x的y次方

格式输出可以把浮点型转换为整型输出并四舍五入

颜色模型转换

  1. #include<stdio.h>
  2. #include<math.h>
  3. #include<stdlib.h>
  4. int max(int R, int G, int B)
  5. {
  6. if (R >= G && R >= B) return R;
  7. if (G >= R && G >= B) return G;
  8. if (B >= R && B >= G) return B;
  9. }
  10. int min(int R, int G, int B)
  11. {
  12. if (R <= G && R <= B) return R;
  13. if (G <= R && G <= B) return G;
  14. if (B <= R && B <= G) return B;
  15. }
  16. int main()
  17. {
  18. int R, G, B;
  19. double V, S, H, n;
  20. scanf("%d%d%d", &R, &G, &B);
  21. V = max(R, G, B);
  22. n = min(R, G, B);
  23. S = (V - n) / V;
  24. if (V == R)
  25. H = 60 * (G - B) / (V - n);
  26. if (V == G)
  27. H = 120 + 60 * (B - R) / (V - n);
  28. if (V == B)
  29. H = 240 + 60*(R - G) / (V - n);
  30. if (V == 0)
  31. S=0;
  32. if (H < 0)
  33. H = H + 360;
  34. printf("%.4lf,%.4lf%%,%.4lf%% ", H, S * 100, V * 100 / 255);
  35. return 0;
  36. }

评论区提供的 

11~20

分数的加、减、乘、除法

  1. #include<stdio.h>
  2. using namespace std;
  3.  
  4. int a_up, a_down, b_up, b_down;
  5. int c_up, c_down;
  6. //分数化简
  7. void simplify(int* up, int* down)
  8. {
  9. 	
  10. 	//化简
  11. 		int i;
  12. 		if (*up > 0)
  13. 		{
  14. 			for (i = *up; i >= 1; i--)
  15. 			{
  16. 				if ((*up % i == 0) && (*down % i == 0))
  17. 				{
  18. 					*up = *up / i;
  19. 					*down = *down / i;
  20. 				}
  21. 			}
  22. 		}
  23. 		if (*up < 0)
  24. 		{
  25. 			for (i = *up; i <=- 1; i++)
  26. 			{
  27. 				if ((*up % i == 0) && (*down % i == 0))
  28. 				{
  29. 					*up = *up / i;
  30. 					*down = *down / i;
  31. 				}
  32. 			}
  33. 		}
  34. 		//使分母恒正，分子决定分数负
  35. 		if (*down < 0)
  36. 		{
  37. 			*up = -*up;
  38. 			*down = -*down;
  39. 		}
  40. }
  41. //输出结果
  42. void output(char c)
  43. {
  44. 	if (c_down == 0)
  45. 		printf("error\n");
  46. 	else if (c_down==1)
  47. 		printf("(%d/%d)%c(%d/%d)=%d\n", a_up, a_down, c, b_up, b_down, c_up);
  48. 	else if(c_up==0)
  49. 		printf("(%d/%d)%c(%d/%d)=0\n", a_up, a_down, c, b_up, b_down);
  50. 	else
  51. 		printf("(%d/%d)%c(%d/%d)=%d/%d\n", a_up, a_down, c, b_up, b_down, c_up, c_down);
  52. }
  53.  
  54. int main()
  55. {
  56. 	//输入分数并化简
  57. 	scanf("%d/%d", &a_up, &a_down);
  58. 	scanf("%d/%d", &b_up, &b_down);
  59. 	simplify(&a_up, &a_down);
  60. 	simplify(&b_up, &b_down);
  61. 	//加法运算并化简输出
  62. 	c_up = a_up * b_down + a_down * b_up;
  63. 	c_down = a_down * b_down;
  64. 	simplify(&c_up, &c_down);
  65. 	output('+');
  66. 	//减法运算并化简输出
  67. 	c_up = a_up * b_down - a_down * b_up;
  68. 	c_down = a_down * b_down;
  69. 	simplify(&c_up, &c_down);
  70. 	output('-');
  71. 	//乘法运算并化简输出
  72. 	c_up = a_up * b_up;
  73. 	c_down = a_down * b_down;
  74. 	simplify(&c_up, &c_down);
  75. 	output('*');
  76. 	//除法运算并化简输出
  77. 	c_up = a_up * b_down;
  78. 	c_down = a_down * b_up;
  79. 	simplify(&c_up, &c_down);
  80. 	output('/');
  81. }

注意审题，输入的数字再输出时要带括号，计算结果不带

注意考虑结果中的整数和等于0的情况

通过for循环时注意正负数寻找最大公因数方式不同，可以分情况或用绝对值解决

定义新函数可以使代码简洁一点

倍数和

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int line,i,t,sum;
  7. 	int n[100000];
  8. 	cin >> line;
  9.  
  10. 	for (i = 1; i <= line; i++)cin >> n[i];
  11. 	for (i = 1; i <= line; i++)
  12. 	{
  13. 		sum = 0;
  14. 		for (t = 3; t <n[i]; t=t+3)
  15. 			sum = sum + t;
  16. 		for (t = 5; t < n[i]; t =t+5)
  17. 			sum = sum + t;
  18. 		for (t = 15; t < n[i]; t=t+15)
  19. 			sum = sum - t;
  20. 		cout << sum << endl;
  21. 	}
  22. 	return 0;
  23. }

动态数组没弄明白，数组直接上取值范围个吧 

幂数模

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	long long a, b, m;
  7. 	long long ans = 1;
  8. 	cin >> a >> b >> m;
  9. 	while (b)
  10. 	{
  11. 		if (b & 1ull)ans = ans* a % m;
  12. 		b >>= 1ull;
  13. 		a = a * a % m;
  14. 	}
  15. 	cout << ans;
  16. }

递推快速幂

级数和

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int n, i;
  7. 	int integer = 1, dec = 2,t=10;
  8. 	double result=0;
  9. 	cin >> n;
  10.  
  11. 	for (i = 1; i <= n; i++)
  12. 	{
  13. 		if (i == 9)
  14. 			t *= 10;
  15. 		if (i == 99)
  16. 			t *= 10;
  17. 		result = result+ (double)integer + (double)dec / t;
  18. 		switch (dec)
  19. 		{
  20. 		case 10:
  21. 			cout << integer << '.' << dec / 10;
  22. 			break;
  23. 		case 100:
  24. 			cout << integer << '.' << dec / 100;
  25. 			break;
  26. 		default:
  27. 				cout << integer << '.' << dec;
  28. 		}
  29. 		if(i<n)
  30. 			cout << '+';
  31. 		else
  32. 			cout << '=' << result;
  33. 		integer++;
  34. 		dec++;
  35. 	}
  36. 	return 0;
  37. }

小数点后无意义的0记得去掉 

组合数

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int a, b, c, d, n, t=0;
  7. 	cin >> n;
  8. 	for (a=0;a<=9;a++)
  9. 	{
  10. 		for (b = 0; b<=9; b++)
  11. 		{
  12. 			for (c = 0; c<=9; c++)
  13. 			{
  14. 				for (d = 0; d<=9; d++)
  15. 				{
  16. 					if ((a + b + c + d) == n)
  17. 						t++;
  18. 				}
  19. 			}
  20. 		}
  21. 	}
  22. 	cout << t;
  23. }

 穷举即可

方阵

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int row, col, i,n;
  7. 	cin >> n;
  8.  
  9. 	for (row = 1; row <= n; row++)
  10. 	{
  11. 		i = row - 1;
  12. 		for (col = 1; col<= n; col++)
  13. 		{
  14. 			cout << i;
  15. 			if (col == n)
  16. 				cout << endl;
  17. 			else
  18. 				cout << ' ';
  19. 			if ((row - col) > 0)
  20. 				i--;
  21. 			if ((row - col) <= 0)
  22. 				i++;
  23.  
  24. 		}
  25. 	}
  26. }

乘数模

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	long long a, b, m, result;
  7. 	cin >> a >> b >> m;
  8.  
  9. 	result = ((a % m) * (b % m)) % m;
  10. 	cout << result;
  11.  
  12. }

直接运算会导致数据溢出，可以根据(a*b)%m=(a%m)*(b%m)%m运算

比率

  1. #include<iostream>
  2. #include<cmath>
  3. using namespace std;
  4. #define PRECISION 0.000001
  5.  
  6. void simplify(int* up, int* down)//化简分数
  7. {
  8.  
  9. 	//化简
  10. 	int i;
  11. 	if (*up > 0)
  12. 	{
  13. 		for (i = *up; i >= 1; i--)
  14. 		{
  15. 			if ((*up % i == 0) && (*down % i == 0))
  16. 			{
  17. 				*up = *up / i;
  18. 				*down = *down / i;
  19. 			}
  20. 		}
  21. 	}
  22. 	if (*up < 0)
  23. 	{
  24. 		for (i = *up; i <= -1; i++)
  25. 		{
  26. 			if ((*up % i == 0) && (*down % i == 0))
  27. 			{
  28. 				*up = *up / i;
  29. 				*down = *down / i;
  30. 			}
  31. 		}
  32. 	}
  33. 	//使分母恒正，分子决定分数负
  34. 	if (*down < 0)
  35. 	{
  36. 		*up = -*up;
  37. 		*down = -*down;
  38. 	}
  39. }
  40.  
  41. int main()
  42. {
  43. 	double n;
  44. 	int up, down;
  45. 	int tens;//用于计算小数部分位数
  46. 	cin >> n;
  47.  
  48. 	double decimal = n - (int)n;//获取小数部分
  49. 	for (tens=1; abs(round(decimal) - decimal) > PRECISION; tens *= 10)
  50. 		decimal *= 10;
  51. 	up = n * tens;
  52. 	down = tens;//把小数乘10的小数位数次并除以十的小数位数次，转化为分数
  53. 	simplify(&up, &down);// 化简
  54.  
  55. 	cout << up << "/" << down;
  56.  
  57. 	return 0;
  58. }

操作数

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. //计算位数
  5. int wide(int a)
  6. {
  7. 	int i;
  8. 	for (i=0;a!=0;i++)
  9. 	{
  10. 		a = a /10;
  11. 	}
  12. 	return i;
  13. }
  14.  
  15. //进行一次减去各位数字和的操作
  16. int operate(int a)
  17. {
  18. 	int times, i, add, res;
  19.  
  20. 	res = a;
  21. 	times = wide(a);
  22. 	for (i = 1; i <= times; i++)
  23. 	{
  24. 		add = a % 10;
  25. 		a = a / 10;
  26. 		res = res - add;//每次循环减去一位数字的值
  27. 	}
  28. 	return res;
  29. }
  30. int main()
  31. {
  32. 	int n, i, add,res;
  33. 	cin >> n;
  34. 	res = n;
  35. 	//计录操作次数
  36. 	for (i = 0; res >0;i++)
  37. 	{
  38. 		res = operate(res);
  39. 	}
  40. 	cout << i;
  41.  
  42. 	return 0;
  43. }

这个写法复杂了，其实可以两个for循环解决的

对称数

  1. ​
  2. #include<iostream>
  3. #include<cmath>
  4. using namespace std;
  5.  
  6. int wide(int a)
  7. {
  8. 	int i;
  9. 	for (i = 0; a != 0; i++)
  10. 	{
  11. 		a = a / 10;
  12. 	}
  13. 	return i;
  14. }
  15.  
  16.  
  17. int main()
  18. {
  19. 	int n, times,i,res;
  20. 	cin >> n;
  21. 	times = wide(n)/2;
  22. 	res = 1;
  23.  
  24. 	for (i = 1; i <= times; i++)
  25. 	{
  26. 		int left =( n / (int)pow(10.0, wide(n)-i))%10;
  27. 		int right = (n % (int)pow(10.0, i)) / (int)pow(10.0, i-1);
  28. 		if (left == 1 && right == 1 || left == 9 && right == 6 || left == 6 && right == 9 || left == 8 && right == 8)
  29. 			res = res;
  30. 		else
  31.  			res = 0;
  32. 	}
  33.  
  34. 	if (res == 1)
  35. 		cout << "Yes";
  36. 		
  37.     else
  38.         cout<<"No";
  39. }
  40.  
  41. ​

 Yes和No别忘了大写

21~30

最大数字

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int N, result, left, right,tens,wrong;
  7. 	cin >> N;
  8.  
  9. 	if (N < 10)cout << N;
  10. 	else left = N / 10 % 10;
  11. 	for (result = N; result >= 10; result--)
  12. 	{
  13. 		
  14. 		for (tens = 10,wrong=0; wrong==0&&result/tens>0; tens *= 10)
  15. 		{
  16. 			left = result / tens % 10;
  17. 			right = (result * 10 / tens) % 10;
  18. 			if (left > right)
  19. 				wrong ++;
  20. 		}
  21. 		if (right == 0) wrong++;
  22. 		if (wrong == 0)
  23. 		{
  24. 			cout << result;
  25. 			break;
  26. 		}
  27. 	}
  28. 	return 0;
  29. }

余数和

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int n, k, sum=0;
  7. 	cin >> n >> k;
  8.  
  9. 	int i;
  10. 	for (i = 1; i <= n; i++)
  11. 	{
  12. 		sum += k % i;
  13. 	}
  14. 	cout << sum;
  15. }

 竖式乘法

  1. #include<iostream>
  2. #include<iomanip>
  3. using namespace std;
  4.  
  5. int widenum(int a)
  6. {
  7. 	int i;
  8. 	for (i = 0; a != 0; i++)
  9. 	{
  10. 		a = a / 10;
  11. 	}
  12. 	return i;
  13. }
  14. int ten(int a)
  15. {
  16. 	int i,ten=1;
  17. 	for (i = 1; i <= a; i++)
  18. 		ten *= 10;
  19. 	return ten;
  20. }
  21.  
  22. int main()
  23. {
  24. 	int n1, n2, wide;
  25. 	int i;
  26. 	cin >> n1 >> n2;
  27. 	wide = widenum(n1 * n2);
  28.  
  29. 	cout << setw(wide+1) << right << n1 << endl;
  30. 	cout << 'x' << setw(wide) << right << n2 << endl;
  31. 	for (i = 0; i <= wide; i++)
  32. 		cout << '-';
  33. 	cout << endl;
  34. 	for (i = 0; i < widenum(n2)-1; i++)
  35. 		cout << setw(wide+1 - i) << right << (n2 / ten(i) % 10) * n1 << endl;
  36. 	cout<<'+'<<setw(wide - i) << right << (n2 / ten(i) % 10) * n1 << endl;
  37. 	for (i = 0; i <= wide; i++)
  38. 		cout << '-';
  39. 	cout << endl;
  40. 	cout <<setw(wide+1)<<right<< n1 * n2;
  41.  
  42. }

阶乘倍数

倒水

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. using namespace std;
  5.  
  6. int Pour(int m, int n, int d) 
  7. {
  8.     int mreal= 0, nreal= 0, step1 = 0, step2 = 0;
  9.     while (mreal != d && nreal != d) {
  10.         if (mreal == 0) {
  11.             mreal = m;
  12.  
  13.         }
  14.         else if (nreal == n) {
  15.             nreal = 0; 
  16.  
  17.         }
  18.         else {
  19.             int total = nreal + mreal > n ? n - nreal : mreal;
  20.             mreal -= total;
  21.             nreal += total;
  22.  
  23.         }
  24.         step1++;
  25.     }
  26.  
  27.     mreal = 0, nreal = 0;
  28.     while (mreal != d && nreal != d) {
  29.         if (nreal == 0) {
  30.             nreal = n;
  31.  
  32.         }
  33.         else if (mreal == m) {
  34.             mreal = 0;
  35.  
  36.         }
  37.         else {
  38.             int total = mreal + nreal > m ? m - mreal : nreal;
  39.             nreal -= total;
  40.             mreal += total;
  41.         }
  42.         step2++;
  43.     }
  44.     int step = step1 > step2 ? step2 : step1;
  45.     return step;
  46. }
  47. int main() {
  48.     int m, n, d;
  49.     cin >> m >> n >> d;
  50.     cout << Pour(m, n, d);
  51.  
  52.     return 0;
  53. }

毕达哥拉斯三元组

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main() {
  5.     int n;
  6.     cin >> n;
  7.     for (int a = 1; a < n; a++) {
  8.         for (int b = a; b < n; b++) {
  9.             int c = n - a - b;
  10.             if (a * a + b * b == c * c) {
  11.                 cout << a * b * c << endl;
  12.                 return 0;
  13.             }
  14.         }
  15.     }
  16.     cout << "No solution" << endl;
  17.     return 0;
  18. }

俄罗斯农夫乘法

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int n1, n2,res=0;
  7. 	cin >> n1 >> n2;
  8.  
  9. 	while (n1 != 0)
  10. 	{
  11. 		if (n1 % 2 != 0)res += n2;
  12. 		cout << n1 << ' ' << n2 << endl;
  13. 		n1 >>=1;
  14. 		n2 <<=1;
  15. 	}
  16. 	cout << res;
  17.  
  18. 	return 0;
  19. }

 查找数列

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int main()
  5. {
  6. 	int i,j, a,n;
  7. 	cin >> n;
  8.  
  9. 	for (i = 1, a = 1; a<n; i++)
  10. 	{
  11. 		for (j = 1; j <= i; j++, a++)
  12. 			if (a == n)break;
  13. 	}
  14. 	cout << j;
  15.  
  16. 	return 0;
  17. }

方案数

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int iter(int n,int i)
  5. {
  6. 	int tmp=0;
  7. 	while (i > 0)
  8. 	{
  9. 		tmp += i;
  10. 		i--;
  11. 		if (tmp == n)return 1;
  12. 		else if(tmp > n)break;
  13. 	}
  14. 	return 0;
  15. }
  16.  
  17. int solution(int n)
  18. {
  19. 	int i, res=1;
  20. 	for (i = (n / 2) + 1; i >= 2; i--)
  21. 		res += iter(n, i);
  22. 	return res;
  23. }
  24.  
  25. int main()
  26. {
  27. 	int n;
  28. 	cin >> n;
  29.  
  30. 	switch (n)
  31. 	{
  32. 	case 1:
  33. 	case 2:
  34. 		cout << '1';
  35. 		break;
  36. 	default:
  37. 		cout << solution(n);
  38. 	}
  39.  
  40. 	return 0;
  41. }

 好数字

  1. #include<iostream>
  2. #include<cmath>
  3. using namespace std;
  4.  
  5. long goodNumber(long N, long i) {
  6.     if (i == N) return 1;
  7.     else if ((i % 2) == 0) {
  8.         return (5 * goodNumber(N, i + 1)) % ((int)pow(10, 9) + 7);
  9.     }
  10.     else if ((i % 2) == 1) {
  11.         return (4 * goodNumber(N, i + 1)) % ((int)pow(10, 9) + 7);
  12.     }
  13. }
  14. int main() 
  15. {
  16.     long N, i = 0;
  17.     cin >> N;
  18.  
  19.     cout << goodNumber(N, i);
  20.  
  21.     return 0;
  22. }

 31~40

可变参数累加

  1. #include<iostream>
  2. #include <initializer_list>//用来实现可变参数
  3. //关于可变参数的实现可以参考https://blog.csdn.net/qq_32534441/article/details/103495144?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522169871084216777224498801%2522%252C%2522scm%2522%253A%252220140713.130102334..%2522%257D&request_id=169871084216777224498801&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~top_positive~default-1-103495144-null-null.142^v96^pc_search_result_base6&utm_term=%E5%8F%AF%E5%8F%98%E5%8F%82%E6%95%B0&spm=1018.2226.3001.4187
  4. // 此处选择 initializer_list标准库类型
  5. //１.initializer_list在C++11中才被引入，这意味着在编译时可能需要加上这个选项 - std = c++11 才能成功编译。上述代码中的auto关键字也是C++11的一部分；
  6. //2. 参数必须放在一组‘{}’（大括号）内，编译器通过大括号来将这组参数转化为initializer_list.大括号的的一组实参与initializer_list形参对应；
  7. //3. 函数原型initializer_list与普通形参无异。这表明形参列表中可以包含其他类型参数且位置不限
  8. using namespace std;
  9.  
  10. int sum(initializer_list<int> il)//函数原型用int实例化initializer_list作为形参
  11. {
  12. 	int sum = 0;
  13. 	for (auto p = il.begin(); p != il.end(); p++) //使用迭代器访问参数
  14. 		sum += *p;
  15. 	return sum;
  16. 	
  17. }
  18.  
  19. int main()
  20. {
  21. 	int a, b, c, d, e, f;
  22. 	cin >> a >> b >> c >> d >> e >> f;
  23.  
  24. 	cout << sum({ a,b,0 }) - sum({ c,d,e,f,0 });//想向initializer_list形参中传递一个值的序列，必须把序列放在一对花括号内
  25.  
  26. 	return 0;
  27. }

哈沙德数 

  1. #include <iostream>
  2. #include <iomanip>
  3. #include <cmath>
  4. using namespace std;
  5. int HarshadNumber(int n){
  6.     int t=n,s=0;
  7.     while(t){
  8.         s=s+t%10;
  9.         t=t/10;
  10.     }
  11.     if(s!=0 && n%s==0) return n/s;
  12.     return 0;
  13. }
  14. int main(){
  15.     int n,i=0;
  16.     cin >> n;
  17.  
  18.     while(n>=2){
  19.         if(HarshadNumber(n)!=0){
  20.             n = HarshadNumber(n);
  21.             i++;
  22.         }
  23.         else break;
  24.     }
  25.     cout << i << endl;
  26.     return 0;
  27. }

幂函数

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. double pow(double base,int exp=2)
  5. {
  6. 	double res=1;
  7. 	for (int i = 1; i <= exp; i++)
  8. 	{
  9. 		res*=base;
  10. 	}
  11. 	return res;
  12. }
  13.  
  14. int main()
  15. {
  16. 	double base;
  17. 	int exp;
  18. 	cin >> base >> exp;
  19.  
  20. 	printf("%.6f\n", pow(base));
  21. 	printf("%.6f", pow(base, exp));
  22.  
  23. 	return 0;
  24. }

素数

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

 光线追踪

  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. }

 基斯数

  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. }

可变参数平均 

  1. #include<iostream>
  2. #include<stdarg.h>
  3. using namespace std;
  4.  
  5. double avg(int n,...)
  6. {
  7. 	int i, val;
  8. 	double sum=0;
  9. 	va_list v1;
  10. 	va_start(v1, n);
  11. 	for (i = 0; i < n; i++)
  12. 	{
  13. 		val = va_arg(v1, double);
  14. 		sum += val;
  15. 	}
  16. 	va_end(v1);
  17. 	return sum / i;
  18. }
  19.  
  20.  
  21. int main()
  22. {
  23. 	double a, b, c, d, e;
  24. 	cin >> a >> b >> c >> d >> e;
  25. 	printf("%.4f", avg(2, a, b) - avg(3, c, d, e));
  26. }

可变参数平均 

  1. #include<iostream>
  2. using namespace std;
  3.  
  4. int PA(int n)
  5. {
  6. 	if (n==0)return 0;
  7. 	else if (n == 1)return 1;
  8. 	return 2 * PA(n - 1) + PA(n - 2);
  9. }
  10.  
  11. int PB(int n)
  12. {
  13. 	int p0 = 0, p1 = 1, pn, i;
  14. 	for (i = 0; i <= n; i++)
  15. 	{
  16. 		if (i == 0)pn = p0;
  17. 		else if (i == 1)pn = p1;
  18. 		else 
  19. 		{
  20. 			pn = 2 * p1 + p0;
  21. 			p0 = p1;
  22. 			p1 = pn;
  23. 		}
  24. 	}
  25. 	return pn;
  26. }
  27.  
  28.  
  29. int main()
  30. {
  31. 	int n;
  32. 	cin >> n;
  33.  
  34. 	if (n % 2 != 0)cout << PA(n);
  35. 	else cout << PB(n);
  36.  
  37. 	return 0;
  38. }

运动会 

  1. 	#include<bits/stdc++.h>
  2. 	using namespace std;
  3. 	const int N = 50000;
  4. 	int vis[N];
  5. 	int prime[N];
  6. 	int phi[N];
  7. 	int sum[N];
  8. 	void get_phi() {
  9. 		phi[1] = 1;
  10. 		int cnt = 0;
  11. 		for (int i = 2; i < N; i++) {
  12. 			if (!vis[i]) {
  13. 				vis[i] = i;
  14. 				prime[cnt++] = i;
  15. 				phi[i] = i - 1;
  16. 			}
  17. 			for (int j = 0; j < cnt; j++) {
  18. 				if (i * prime[j] > N) break;
  19. 				vis[i * prime[j]] = prime[j];
  20. 				int m = i * prime[j];
  21. 				if (i % prime[j] == 0) { // m = prime[i] * i * ````
  22. 					phi[m] = phi[i] * prime[j]; 
  23. 					break;
  24. 				}
  25. 				phi[m] = phi[i] * phi[prime[j]];
  26. 			}
  27. 		}
  28. 	}
  29. 	int main() {
  30. 		get_phi();
  31. 		sum[1] = 1;
  32. 		for (int i = 2; i <= N; i++)
  33. 			sum[i] = sum[i - 1] + phi[i];
  34. 		int n;
  35. 		cin >> n;
  36. 		if (n == 1) cout << "0" << endl;
  37. 		else {
  38. 			printf("%d\n", 2 * sum[n - 1] + 1);
  39. 		}
  40. 		return 0;
  41. 	}

欧拉筛法解决

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