LightOJ 1231 Coin Change (I)(部分背包)

B - Coin Change (I)

Time Limit:1000MS    Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1231

Description

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 valueA₁, A₂ ... A_n respectively. You have to find the number of ways you can makeK 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 ifK = 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) andK (1 ≤ K ≤ 1000). The next line contains 2n integers, denotingA₁, A₂ ... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 100, 1 ≤ C_i ≤ 20). AllA_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 modulo100000007.

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

  对于dp的思想还是不懂，对于背包问题更是理解不透彻，既然发现了这一点就要赶紧完善自己！！

这是背包题目要对物品种类，背包数量，物品的个数一一枚举，根据剩余的背包空间进行下一步

思路：dp[i][j]+=dp[i-1][j-k*a[i]];

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[55][1005];//根据k的范围定义dp的数组
int main()
{
    int t,n,k,a[55],b[55],m=1;
    cin>>t;
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
            for(int i=1;i<=n;i++)
                scanf("%d",&b[i]);
                //
            memset(dp,0,sizeof(dp));
            dp[0][0]=1;
            //控制输出格式
         if(m++)
            cout<<"Case "<<m-1<<": ";

            for(int i=1;i<=n;i++)//枚举物品种类
            {
                for(int j=0;j<=k;j++)//枚举背包数量
                    for(int k=1;k<=b[i];k++)//枚举物品的个数
                        if(j-k*a[i]>=0)//判断背包当前是否可以继续放物品
                            dp[i][j]+=dp[i-1][j-k*a[i]];
                for(int h=0;h<=k;h++)
                    dp[i][h]=(dp[i][h]+dp[i-1][h])%100000007;
            }
            printf("%d\n",dp[n][k]);
    }
    return 0;
}