oyoj-43 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

24点游戏，牌的数量不定，点数之和也不定。先从数组任取两个数，四则运算这两个数，将结果放回数组，这样就变成就n-1的子问题了。递归实现。

 #include <iostream> #include <stdio.h>#include <cmath>#include <cstring>#define eps 10E-6using namespace std;int n;double m;double a[110];int dfs(int num){if(num==n){if(abs(a[n]-m)<=eps) //跳出条件 不懂  可是换个就会错  。。。。//判断若a[n]与m是否无限接近的吧，则返回1； return 1;return 0;}for(int i=num;i<n;i++) {for(int j=i+1;j<=n;j++)       {double p,q;q=a[i];p=a[j];a[i]=a[num];  //将值赋值给数组 //遍历所有可能 a[j]=q+p;  if(dfs(num+1))return 1;a[j]=q-p;  //减法1 if(dfs(num+1))return 1;a[j]=p-q;  //减法2 if(dfs(num+1))return 1;a[j]=p*q;if(dfs(num+1))return 1;if(q)  //q!=0a[j]=p/q;  if(dfs(num+1))return 1;if(p) //p!=0 a[j]=q/p;if(dfs(num+1))return 1;a[i]=q;  a[j]=p;}}return 0;}int main(){int t;cin>>t;while(t--){cin>>n>>m;for(int i=1;i<=n;++i){cin>>a[i];}if(dfs(1))cout<<"Yes"<<endl;elsecout<<"No"<<endl;}return 0;}