(1231)lightOJ

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. C₁, C₂, ... C_n denote the number of coins of value A₁, A₂ ... A_n respectively. You have to find the number of ways you can make K using the coins.

For example, suppose there are three coins 1, 2, 5 and we can use coin 1 at most 3 times, coin 2 at most 2 times and coin 5 at most 1 time. Then if K = 5 the possible ways are:

1112

122

5

So, 5 can be made in 3 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 ≤ 50) and K (1 ≤ K ≤ 1000). The next line contains2n integers, denoting A₁, A₂ ... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 100, 1 ≤ C_i ≤ 20). 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 3 2 1

4 20

1 2 3 4 8 4 2 1

Sample Output

Case 1: 3

Case 2: 9

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>




#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define eps 1e-8
#define INF 0xfffffff
#define mod  100000007
using namespace std;


int w[MAX];
int c[MAX];
int v;
int n;

/*

题意：给你不同种类的硬币和个数，让你求组成面值为k的方法数

解法：最开始觉得是多重背包，然后一顿找模板，最后发现只是个简单DP。还有啊，对多重背包那里实在是搞得不好，不扎实，昨天打的好渣的说。

*/
int dp[55][1009];//前i种物品组成容量为j的取法


int solve_dp()
{
    memset(dp,0,sizeof(dp));
    for(int i = 1;i<=n;i++)
    {
        for(int j = 1;j<=c[i];j++)
        {
            if(j*w[i]<=v)//小于预期额度
            dp[i][j*w[i]] = 1;//选择前i种硬币下可以凑成的额度
        }
    }
    for(int i = 2;i<=n;i++)
    {
        for(int j = 1;j<=v;j++)
        {
            for(int l = 0;l<=c[i];l++)
            {
                if(j-l*w[i]>0)
                dp[i][j] = (dp[i][j] + dp[i-1][j - l*w[i]]) %mod;//当前的方法数+前i-1种硬币组成j - l*w[i]的额度的方法数
            }
        }
    }
    return dp[n][v];
}
int main()
{
    int d = 1;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&v);
        for(int i = 1;i<=n;i++) scanf("%d",&w[i]);
        for(int i = 1;i<=n;i++) scanf("%d",&c[i]);
        printf("Case %d: %d\n",d++,solve_dp());
    }


    return 0;
}