C++计算24点的简单程序分享！

本文实例为大家分享了C++计算24点的的具体代码，供大家参考，具体内容如下

近来家庭作业里有24点的题目，为了找出所有可能的组合，就写了个简单的程序：

1. 运行程序

2. 输入4个整数,比如：3  3  7  8

3. 显示所有可能的组合

代码：

  #include "assert.h"  #include <iostream>     double operate(double num1, double num2, int op)  {   assert(op >= 0 && op < 4);      if(op == 0){   return num1 + num2;   }   else if(op == 1){   return num1 - num2;   }   else if(op == 2){   return num1 * num2;   }   else{   return num1/num2;   }  }     int calculate(int num1, int num2, int num3, int num4)  {   char operators[] = "+-*/";      for(int i = 0; i < 4; i ++)   {   for(int j = 0; j < 4; j ++)   {    for (int k = 0; k < 4; k ++)    {    double ret = operate(num1, num2, i);    ret = operate(ret, num3, j);    ret = operate(ret, num4, k);    if(abs(ret - 24) < 0.001){     printf("((%d %c %d) %c %d) %c %d = %fn", num1, operators[i],                       num2, operators[j],         num3, operators[k],         num4, ret);    }       ret = operate(num1, num2, i);    double ret2 = operate(num3, num4, k);    ret = operate(ret, ret2, j);    if(abs(ret - 24) < 0.001){     printf("(%d %c %d) %c (%d %c %d) = %fn", num1, operators[i],                       num2, operators[j],         num3, operators[k],         num4, ret);    }       ret = operate(num2, num3, j);    ret = operate(num1, ret, i);    ret = operate(ret, num4, k);    if(abs(ret - 24) < 0.001){     printf("(%d %c (%d %c %d)) %c %d = %fn", num1, operators[i],                       num2, operators[j],         num3, operators[k],         num4, ret);    }       ret = operate(num2, num3, j);    ret = operate(ret, num4, k);    ret = operate(num1, ret, i);    if(abs(ret - 24) < 0.001){     printf("%d %c ((%d %c %d) %c %d) = %fn", num1, operators[i],                      num2, operators[j],         num3, operators[k],         num4, ret);    }       ret = operate(num3, num4, k);    ret = operate(num2, ret, j);    ret = operate(num1, ret, i);    if(abs(ret - 24) < 0.001){     printf("%d %c (%d %c (%d %c %d)) = %fn", num1, operators[i],                       num2, operators[j],         num3, operators[k],         num4, ret);    }    }   }   }   return 0;  }     int main(int argc, char* argv[])  {      int nums[4] = {0, 0, 0, 0};   std::cin >> nums[0] >> nums[1] >> nums[2] >> nums[3];      for (int i = 0; i < sizeof(nums)/sizeof(nums[0]); i ++)   {   int num1 = nums[i];   int ret = num1;       for(int j = 0; j < sizeof(nums)/sizeof(nums[0]); j ++)   {    if(j == i)    continue;       int num2 = nums[j];        for(int k = 0; k < sizeof(nums)/sizeof(nums[0]); k++)    {    if( k == i || k == j)     continue;       int num3 = nums[k];       for(int l = 0; l < sizeof(nums)/sizeof(nums[0]); l ++)    {     if(l == i || l == j || l == k)     continue;        int num4 = nums[l];     calculate(num1, num2, num3, num4);    }    }   }   }      return 0;  }

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