nyoj 43 24 Point game（dfs暴力）

 

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

深搜枚举各种可能。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1000000
#define inf 1e12
int n;
double v,a[6];
bool dfs(int t){
   if(t==n){
      if(fabs(v-a[n])<1e-7){
         return true;
      }
      return false;
   }
   for(int i=t;i<n;i++){
      for(int j=i+1;j<=n;j++){
         double temp1=a[i];
         double temp2=a[j];
         a[i]=a[t];
         a[j]=temp1+temp2; if(dfs(t+1)) return true;//加
         a[j]=temp1-temp2; if(dfs(t+1)) return true;//减1
         a[j]=temp2-temp1; if(dfs(t+1)) return true;//减2
         a[j]=temp1*temp2; if(dfs(t+1)) return true;//乘
         if(temp1){//除1
            a[j]=temp2/temp1; if(dfs(t+1)) return true;
         }
         if(temp2){//除2
            a[j]=temp1/temp2; if(dfs(t+1)) return true;
         }
         a[i]=temp1;//还原
         a[j]=temp2;//还原
      }
   }
   return false;
}
int main()
{
   int t;
   scanf("%d",&t);
   while(t--){
      scanf("%d%lf",&n,&v);
      for(int i=1;i<=n;i++){
         scanf("%lf",&a[i]);
      }
      if(dfs(1)){
         printf("Yes
");
      }else{
         printf("No
");
      }
   }
    return 0;
}