24点游戏游戏(C++)

最新推荐文章于 2022-12-10 15:43:38 发布

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最新推荐文章于 2022-12-10 15:43:38 发布

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分类专栏：作业文章标签： c++

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本文介绍了如何使用C++编程实现经典的24点游戏算法。通过解析给定的四个数字，运用加减乘除及括号操作，判断是否能得出24。内容涵盖了数字组合、算术运算及递归思想的应用。

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#include<bits/stdc++.h>
using namespace std;
int card[] = { 1,2,3,4,5,6,7,8,9,10,11,12,13 };
int nums[4];
char ope[4] = { '+','-','*','/' };
 
 
double calcute(double a, double b, char index)
{
	if (index == '+') return a + b;  
	else if (index == '-') return a - b;
	else if (index == '*') return a*b;
	else if (index == '/')
		if (b != 0)
			return a / b;
}


int main()
{
	int i = 0;
	int j;
	cout << "生成的四张牌面为：";
	srand((unsigned)time(0));
	for (i = 0; i<4; i++)
	{
	     j =rand() % 13 ;
		nums[i] = card[j];
	}
	cout << nums[0] << ";" << nums[1] << ";" << nums[2] << ";" << nums[3] << "。" << endl;
	double  temp[3], tem[2];
	double  sum; 
	int  judge = 0;
	for (int i = 0; i < 4; i++)
	{
		for (int j = 0; j < 4; j++)
		{
			for (int k = 0; k < 4; k++)  
			{
				for (int m = 0; m < 3; m++)
				{
					if (nums[m + 1] == 0 && ope[i] == '/') break;
					temp[m] = calcute(nums[m], nums[m + 1], ope[i]);
					temp[(m + 1) % 3] = nums[(m + 2) % 4];
					temp[(m + 2) % 3] = nums[(m + 3) % 4];     
					for (int n = 0; n < 2; n++)
					{
						if (temp[n + 1] == 0 && ope[j] == '/') break;
						tem[n] = calcute(temp[n], temp[n + 1], ope[j]);
						tem[(n + 1) % 2] = temp[(n + 2) % 3];    
						if (tem[1] == 0 && ope[k] == '/') break;
						sum = calcute(tem[0], tem[1], ope[k]);             
						if (sum == 24)    
						{
							judge = 1;       
							if (m == 0 && n == 0)
								cout << "((" << nums[0] << ope[i] << nums[1] << ")" << ope[j] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
							else if (m == 0 && n == 1)
								cout << "(" << nums[0] << ope[i] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[j] << nums[3] << ")=" << sum << endl;
							else if (m == 1 && n == 0)
								cout << "(" << nums[0] << ope[j] << "(" << nums[1] << ope[i] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
							else if (m == 1 && n == 1)
								cout << nums[0] << ope[k] << "((" << nums[1] << ope[i] << nums[2] << ")" << ope[j] << nums[3] << ")=" << sum << endl;
							else if (m == 2 && n == 0)
								cout << "(" << nums[0] << ope[j] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[i] << nums[3] << ")=" << sum << endl;
							else if (m == 2 && n == 0)
								cout << nums[0] << ope[k] << "(" << nums[1] << ope[j] << "(" << nums[2] << ope[i] << nums[3] << "))=" << sum << endl;
						}
					}
				}
			}
		}
	}
	if (judge == 0)
		cout << "这四张扑克牌无法找到一个合理的解" << endl;
	return 0;
}