24 Point game

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。

24 24 3 3 8 83 24 8 3 3樣例輸出
YesNo來源經(jīng)典改編
//一開始只有起始的幾個數(shù) ，然后通過回溯每次取任意兩個數(shù)進行加、減、乘、除、被減、被除6種操作，然后把結(jié)果放到集合中，并且在集合中刪去之前操作的兩個數(shù)，直到集合中只有一個數(shù)了，那么比較一下是不是24（注意因為過程中有除法的存在，這里就涉及到一個精度問題，多次運算過后可能會出現(xiàn)24.000001或者23.999999這種情況，所以要注意，不能單純的（int）一下
#include <bits/stdc++.h>using namespace std;bool vis[15];double tar,arr[15];int n;double judge(double x,double y,int k){    switch(k)    {        case 0: return x + y;        case 1: return x - y;        case 2: return x * y;        case 3: return x / y;        case 4: return y - x;        case 5: return y / x;    }}bool dfs(int num,int p){    if(num == 1 && fabs(arr[p] - tar)< 1e-6)        return true;    double temp_1,temp_2;    for(int i=0;i<n;i++)    {        if(!vis[i])            for(int j=i+1;j<n;j++)        {            if(!vis[j])            {                temp_1 = arr[i];                temp_2 = arr[j];                for(int k=0;k<6;k++)                {                    if(k == 3&&arr[j] == 0||k == 5&&arr[i] == 0)                        continue;                    arr[j] = judge(arr[i],arr[j],k);                    vis[i] = true;                    if(dfs(num-1,j))                        return true;                    vis[i] = false;                    arr[i] = temp_1;                    arr[j] = temp_2;                }            }        }    }    return false;}int main(){    int Case;    cin>>Case;    while(Case--)    {       cin>>n>>tar;       for(int i=0;i<n;i++)       {           cin>>arr[i];       }       for(int i=0;i<15;i++)        vis[i] = false;        if(dfs(n,n-1))            cout<<"Yes"<<endl;        else cout<<"No"<<endl;    }    return 0;}