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[],c[],dp[],b[];
 int main()
 {
     int n,k,t,i,j,r;
     scanf("%d",&t);
     for(i=; i<=t; i++)
     {
         memset(dp,,sizeof(dp));
         scanf("%d%d",&n,&k);
         for(j=; j<=n; j++)scanf("%d",&b[j]);
         for(j=; j<=n; j++)scanf("%d",&c[j]);
         int nu=,re,rs;
         for(j=; 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>>=;
                 re<<=;
             }
         }
         for(j=; j<nu; j++)
         {
             for(r=k; r>=; r--)
             {
                 if(dp[r]&&r+a[j]<=k)
                     dp[r+a[j]]=;
                 if(r==&&a[j]<=k)
                     dp[a[j]]=;
             }

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

 }