nyoj43 24 Point game(搜索)

最新推荐文章于 2017-11-06 20:07:44 发布

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#nyoj43 24 Point game

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本文介绍了一种经典的数学游戏——24点游戏的算法实现。玩家需要利用给定的几个数字通过加减乘除运算得到24。文章详细解释了如何通过递归和枚举的方式，验证一组数字是否能通过规定的运算达到目标值。

24 Point game

时间限制：3000 ms | 内存限制：65535 KB

难度：5

描述

There is a game which is called 24 Point game.

In this game , you will be given some numbers. Your task is to find an expression which have all the given numbers and the value of the expression should be 24 .The expression mustn't have any other operator except plus,minus,multiply,divide and the brackets.

e.g. If the numbers you are given is "3 3 8 8", you can give "8/(3-8/3)" as an answer. All the numbers should be used and the bracktes can be nested.

Your task in this problem is only to judge whether the given numbers can be used to find a expression whose value is the given number。

输入

The input has multicases and each case contains one line
The first line of the input is an non-negative integer C(C<=100),which indicates the number of the cases.
Each line has some integers,the first integer M(0<=M<=5) is the total number of the given numbers to consist the expression,the second integers N(0<=N<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers is non-negative and less than 100

输出

For each test-cases,output "Yes" if there is an expression which fit all the demands,otherwise output "No" instead.

样例输入

2
4 24 3 3 8 8
3 24 8 3 3

样例输出

Yes
No

题意：24点，给你n个数和一个

值，如果这n个数通过+，-，*，/，（）组合等于V，则输出Yes，否则No

//搜索：枚举N个数进行+-/*运算将运算的结果保存在数组里即可模拟枚举出（）运算
#include<stdio.h>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
double r;//点数 
double mat[5];
int m;//个数 
int bfs(int k){
	if(k==m-1){
		if(fabs(mat[m-1]-r)<1e-6)//判断是否相等 
		return 1;
		else return 0;
	}
	for(int i=k;i<m-1;i++){
		for(int j=i+1;j<m;j++){
			double a=mat[i],b=mat[j];
			mat[i]=mat[k];
			mat[j]=a+b;if(bfs(k+1)) return 1;
			mat[j]=a-b;if(bfs(k+1)) return 1;
			mat[j]=a*b;if(bfs(k+1)) return 1;
			mat[j]=b-a;if(bfs(k+1)) return 1;
			mat[j]=b+a;if(bfs(k+1)) return 1;
			if(b!=0){
				mat[j]=a/b;if(bfs(k+1)) return 1;
			}
			if(a!=0){
				mat[j]=b/a;if(bfs(k+1)) return 1;
			}
			mat[i]=a;mat[j]=b;
		}
	}
	return 0;
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--){
		scanf("%d%lf",&m,&r);
		for(int i=0;i<m;i++){
			scanf("%lf",&mat[i]);
		}
	if(bfs(0))
	printf("Yes\n");
	else printf("No\n");	
	}
	return 0;
}