24点游戏

最新推荐文章于 2025-03-10 12:30:00 发布

转载最新推荐文章于 2025-03-10 12:30:00 发布 · 1k 阅读

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#24点 #递归

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本文介绍了一个基于递归思想实现的24点游戏求解算法，通过枚举所有可能的操作组合来寻找达到目标值24的方法。该程序使用C++编写，能够在控制台输入四个数字后输出可行的数学表达式。

// code.cpp : 定义控制台应用程序的入口点。    
// editor: Visual C++ 2010 Ultimate    
// system: win7 x32
// create: 2013-08-02 10:46AM    
// author: starfish(海星)
// remark: 转载自大神的文章片段，提交pongo需修改

#include "stdafx.h"
#include <iostream>
#include <string>
#include <cmath>
using namespace std;

const double PRECISION = 1E-6;
const int COUNT_OF_NUMBER = 4;
const int NUMBER_TO_CAL = 24;

double number[COUNT_OF_NUMBER];
string expression[COUNT_OF_NUMBER];

bool Search(int n);

int _tmain(int argc, _TCHAR* argv[])
{
	for(int i=0;i<COUNT_OF_NUMBER; i++){
		char buffer[20];
		int x;
		cin>>x;
		number[i] = x;
		itoa(x, buffer, 10);
		expression[i] = buffer;
	}

	if(Search(COUNT_OF_NUMBER)){
		cout<< "SUCCESS" <<endl;
	} else {
		cout<< "FAILURE" <<endl;
	}
	system("pause");
	return 0;
}

bool Search(int n)
{
	if(n==1){
		if(fabs(number[0]-NUMBER_TO_CAL)<PRECISION){
			cout<<expression[0]<<endl;
			return true;
		} else {
			return false;
		}
	}
	for(int i=0; i<n; i++){
		for(int j=i+1; j<n; j++){
			double a,b;
			string expa, expb;

			a = number[i];
			b = number[j];
			number[j] = number[n-1];

			expa = expression[i];
			expb = expression[j];
			expression[j] = expression[n-1];
			
			expression[i] = '(' + expa + '+' + expb + ')';
			number[i] = a + b;
			if(Search(n-1)) return true;
			
			expression[i] = '(' + expa + '-' + expb + ')';
			number[i] = a - b;
			if(Search(n-1)) return true;

			expression[i] = '(' + expb + '-' + expa + ')';
			number[i] = b - a;
			if(Search(n-1)) return true;

			expression[i] = '(' + expa + '*' + expb + ')';
			number[i] = a * b;
			if(Search(n-1)) return true;

			if(b!=0){
				expression[i] = '(' + expa + '/' + expb + ')';
				number[i] = a / b;
				if(Search(n-1)) return true;
				//-- 算分数24点 jkj --//
				expression[i] = '(' + expb + '/' + expa + ')';
				number[i] = b / a;
				if(Search(n-1)) return true;
			}

			number[i] = a;
			number[j] = b;
			expression[i] = expa;
			expression[j] = expb;

		}
	}
	return false;
}