| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
Given n coins, values of them are A1, A2 ... An 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
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <algorithm> 6 using namespace std; 7 int a[50]; 8 bool dfs(int n,int re,int sh) 9 { 10 if(re==0||sh==re)return true; 11 if(n==0||re<0||re>sh)return false; 12 if(dfs(n-1,re,sh-2*a[n]))return true; 13 if(dfs(n-1,re-a[n],sh-2*a[n]))return true; 14 if(dfs(n-1,re-2*a[n],sh-2*a[n]))return true; 15 return false; 16 } 17 int main() 18 { 19 int t,n,k,i,j,sum; 20 scanf("%d",&t); 21 for(i=1; i<=t; i++) 22 { 23 sum=0; 24 scanf("%d%d",&n,&k); 25 for(j=1; j<=n; j++)scanf("%d",&a[j]); 26 sort(a+1,a+n+1); 27 for(j=1; j<=n; j++)sum+=a[j]<<1; 28 cout<<"Case "<<i<<": "; 29 if(dfs(n,k,sum)) 30 cout<<"Yes"<<endl; 31 else cout<<"No"<<endl; 32 } 33 }
Coin Change (IV) （dfs）,布布扣,bubuko.com
原文：http://www.cnblogs.com/ERKE/p/3582222.html
