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 = %f\n", 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) = %f\n", 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 = %f\n", 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) = %f\n", 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)) = %f\n", 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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