Light oj 1231 - Coin Change (I)(dp)

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本文探讨了一种典型的组合计数问题——硬币组合问题。在限定每种面额硬币使用次数的情况下，如何计算组成特定金额的方法总数。通过动态规划算法解决该问题，并给出具体的实现代码。

1231 - Coin Change (I)

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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

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 contains 2n 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

Output for Sample Input

3 5

1 2 5 3 2 1

4 20

1 2 3 4 8 4 2 1

Case 1: 3

Case 2: 9

PROBLEM SETTER: JANE ALAM JAN

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-12

typedef long long ll;
using namespace std;
#define mod  100000007
#define N 1005

int n,a[N],num[N];
int dp[55][N];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/ksh/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t,ca=0;
    scanf("%d",&t);
    int all;
    while(t--)
    {
        scanf("%d%d",&n,&all);

        for(int i=1;i<=n;i++)
             scanf("%d",&a[i]);

        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;

        for(int i=1;i<=n;i++)
           for(int v=0;v<=all;v++)
           {
               int t=v/a[i];
               for(j=0;j<=t&&j<=num[i];j++)
                  dp[i][v]=(dp[i][v]+dp[i-1][v-j*a[i]])%mod;
           }

       printf("Case %d: %d\n",++ca,dp[n][all]);

    }
    return 0;
}