Coin Change (II)(完全背包)

                                                           Coin Change (II)

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. You have to find the number of ways you can make K using the coins. You can use any coin at most K times.

For example, suppose there are three coins 1, 2, 5. Then if K = 5 the possible ways are:

11111

1112

122

5

So, 5 can be made in 4 ways.

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 ≤ 100) and K (1 ≤ K ≤ 10000). The next line contains n integers, denoting A₁, A₂ ... A_n (1 ≤ A_i ≤ 500). All A_i will be distinct.

Output

For each case, print the case number and the number of ways K can be made. Result can be large, so, print the result modulo 100000007.

Sample Input

2

3 5

1 2 5

4 20

1 2 3 4

Sample Output

Case 1: 4

Case 2: 108

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 #define mod 100000007
 int a[],c[],dp[];
 int main()
 {
     int n,k,t,i,j,r,w;
     scanf("%d",&t);
     for(i=; i<=t; i++)
     {
         memset(dp,,sizeof(dp));
         scanf("%d%d",&n,&k);
         for(j=; j<=n; j++)scanf("%d",&a[j]);
         dp[]=;
         for(j=; j<=n; j++)
         {
             for(r=; r<=k; r++)
             {
                 if(r+a[j]<=k)
                 {
                     dp[r+a[j]]+=dp[r],dp[r+a[j]]%=mod;
                 }
             }
         }
         cout<<"Case "<<i<<": ";
         cout<<dp[k]<<endl;
     }
 }