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：
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#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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以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/cjzs1108/article/details/40628431?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task
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