C++计算24点的简单程序

 更新时间:2020年03月06日 13:30:59   作者:icedump  
这篇文章主要为大家详细介绍了C++计算24点的简单程序,文中示例代码介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们可以参考一下

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;
}

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脚本之家。

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