C++计算24点的简单程序

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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