C - Coin Change (III)（多重背包 二进制优化）

C - Coin Change (III)

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1233

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 value A₁, A₂ ... A_nrespectively. You have to find the number of different values (from 1 to m), which can be produced using these coins.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 100), m (0 ≤ m ≤ 10⁵). The next line contains 2n integers, denoting A₁, A₂... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 10⁵, 1 ≤ C_i ≤ 1000). All A_i will be distinct.

Output

For each case, print the case number and the result.

Sample Input

2

3 10

1 2 4 2 1 1

2 5

1 4 2 1

Sample Output

Case 1: 8

Case 2: 4

 

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define mod 100000007
int a[1001],c[101],dp[100001],b[101];
int main()
{
    int n,k,t,i,j,r;
    scanf("%d",&t);
    for(i=1; i<=t; i++)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&k);
        for(j=1; j<=n; j++)scanf("%d",&b[j]);
        for(j=1; j<=n; j++)scanf("%d",&c[j]);
        int nu=0,re,rs;
        for(j=1; j<=n; j++)
        {
            re=b[j],rs=c[j];
            c[j]*=b[j];
            while(rs)
            {
                if(re<c[j])
                    a[nu++]=re,c[j]-=re;
                else a[nu++]=c[j];
                rs>>=1;
                re<<=1;
            }
        }
        for(j=0; j<nu; j++)
        {
            for(r=k; r>=0; r--)
            {
                if(dp[r]&&r+a[j]<=k)
                    dp[r+a[j]]=1;
                if(r==0&&a[j]<=k)
                    dp[a[j]]=1;
            }

        }
        int sum=0;
        for(r=0; r<=k; r++)sum+=dp[r];
        cout<<"Case "<<i<<": ";
        cout<<sum<<endl;
    }

}