Coin Change (IV) （dfs）

Coin Change (IV)

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

Given n coins, values of them are A₁, A₂ ... A_n respectively, you have to find whether you can pay K using the coins. You can use any coin at most two times.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 18) and K (1 ≤ K ≤ 10⁹). The next line contains n distinct integers denoting the values of the coins. These values will lie in the range [1, 10⁷].

Output

For each case, print the case number and 'Yes' if you can pay K using the coins, or 'No' if it's not possible.

Sample Input

3

2 5

1 2

2 10

1 2

3 10

1 3 5

Sample Output

Case 1: Yes

Case 2: No

Case 3: Yes

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 int a[];
 bool dfs(int n,int re,int sh)
 {
     if(re==||sh==re)return true;
     if(n==||re<||re>sh)return false;
     if(dfs(n-,re,sh-*a[n]))return true;
     if(dfs(n-,re-a[n],sh-*a[n]))return true;
     if(dfs(n-,re-*a[n],sh-*a[n]))return true;
     return false;
 }
 int main()
 {
     int t,n,k,i,j,sum;
     scanf("%d",&t);
     for(i=; i<=t; i++)
     {
         sum=;
         scanf("%d%d",&n,&k);
         for(j=; j<=n; j++)scanf("%d",&a[j]);
         sort(a+,a+n+);
         for(j=; j<=n; j++)sum+=a[j]<<;
         cout<<"Case "<<i<<": ";
         if(dfs(n,k,sum))
             cout<<"Yes"<<endl;
         else cout<<"No"<<endl;
     }
 }