Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
In a strange shop there are n types of coins of value A1, A2 ... An. 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 ... An (1 ≤ Ai ≤ 500). All Ai 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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 #define mod 100000007 6 int a[11000],c[10000],dp[11000]; 7 int main() 8 { 9 int n,k,t,i,j,r,w; 10 scanf("%d",&t); 11 for(i=1; i<=t; i++) 12 { 13 memset(dp,0,sizeof(dp)); 14 scanf("%d%d",&n,&k); 15 for(j=1; j<=n; j++)scanf("%d",&a[j]); 16 dp[0]=1; 17 for(j=1; j<=n; j++) 18 { 19 for(r=0; r<=k; r++) 20 { 21 if(r+a[j]<=k) 22 { 23 dp[r+a[j]]+=dp[r],dp[r+a[j]]%=mod; 24 } 25 } 26 } 27 cout<<"Case "<<i<<": "; 28 cout<<dp[k]<<endl; 29 } 30 }