24 Point game

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

思路：

数组的精巧利用， 既节约时间也节约空间， 而且还可以不用处理括号。

#include 
#include 

int n;
double val, num[10];

int dfs(int cnt)
{
	int i, j;
	double left, right;
	if(cnt >= n)
	{
		if(fabs(num[n] - val) < 0.000001)     //因double有误差
		{
			return 1;
		}
		else
		{
			return 0;
		}
	}
	for(i = cnt; i < n; i++)                 //仔细理解此处妙处
	{
		for(j = i+1; j <= n; j++)
		{
			left = num[i];                 //left,right是保存数据便于回溯用的， 还有运算时也要用
			right = num[j];
			num[i] = num[cnt];             //在i>cnt时， 使num[i]存的是未处理数据
			num[j] = left+right;           //6种运算结果存入num[j]中， 存新数据（减和除存两次）
			if(dfs(cnt+1))
			{
				return 1;
			}
			num[j] = left-right;
			if(dfs(cnt+1))
			{
				return 1;
			}
			num[j] = right-left;
			if(dfs(cnt+1))
			{
				return 1;
			}
			num[j] = left*right;
			if(dfs(cnt+1))
			{
				return 1;
			}
			if(left)                    //注意除0
			{
				num[j] = right/left;
				if(dfs(cnt+1))
				{
					return 1;
				}
			}
			if(right)
			{
				num[j] = left/right;
				if(dfs(cnt+1))
				{
					return 1;
				}
			}
			num[i] = left;                   //回溯还原场景
			num[j] = right;
		}
	}
	return 0;
}

int main()
{
	int t, i;
	scanf("%d", &t);
	while(t--)
	{
		scanf("%d%lf", &n, &val);
		for(i = 1; i <= n; i++)
		{
			scanf("%lf", &num[i]);
		}
		if(dfs(1))
		{
			printf("Yes\n");
		}
		else
		{
			printf("No\n");
		}
	}
	return 0;
}