深搜 nyoj 43 24 Point game

24 Point game

時間限制：3000 ms  |  內存限制：65535 KB

難度：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。

輸入

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

輸出

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

樣例輸入

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

樣例輸出

Yes
No

來源

經典改編

上傳者

張雲聰

題目不難，只有四種可能，逐個遍歷搜索就行！

代碼如下：

//考查知識點：深搜
//因爲全角，半角的輸入問題，看了一個小時。。。 
#include<stdio.h>
#include<math.h>
#include<string.h>
#define eps 10E-6
int n;
double sum;
double a[110];
int dfs(int num)
{
	if(num==n)
	{
		if(fabs(a[n]-sum)<=eps)//跳出條件 
		return 1;
		return 0;
	}
	for(int i=num;i<n;++i)//從num開始保證了深度遍歷的時候，數字逐步後移 
	{
		for(int j=i+1;j<=n;++j)
		{
			double right,left;
			left=a[i];
			right=a[j];
			a[i]=a[num];//將值賦值給數組 
			
			a[j]=left+right;//遍歷每一種可能的操作 
			if(dfs(num+1))
			return 1;
			
			a[j]=left-right;//大小不確定 
			if(dfs(num+1))
			return 1;
			
			a[j]=-left+right;
			if(dfs(num+1))
			return 1;
			
			a[j]=left*right;
			if(dfs(num+1))
			return 1;
			
			if(right)
			a[j]=left/right;//除數不爲零 
			if(dfs(num+1))
			return 1;
			
			if(left)
			a[j]=right/left;
			if(dfs(num+1))
			return 1;
			
			a[i]=left;//回溯 
			a[j]=right;
		}
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%lf",&n,&sum);
		for(int i=1;i<=n;++i)
		{
			scanf("%lf",&a[i]);
		}
		if(dfs(1))
		puts("Yes");
		else
		puts("No");
	}
	return 0;
}