（多重背包）Coins

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18021    Accepted Submission(s): 7085

Problem Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

 

Input

The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

 

Output

For each test case output the answer on a single line.

 

Sample Input

3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0

 

Sample Output

8 4

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

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解题：多重背包的一个题目。

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
#define maxn 110
#define maxm 100010
int a[maxn],c[maxn],vis[maxm],num[maxm]; //vis[j]标记，标记能否刚好装满重量为j的物品

int main()
{
    int n,m;
    while(cin>>n>>m){
        if(n==0&&m==0)
            break;
        for(int i=0;i<n;i++)
            cin>>a[i];
        for(int i=0;i<n;i++)
            cin>>c[i];
        memset(vis,0,sizeof(vis));
        vis[0]=1;
        int cnt=0;
        for(int i=0;i<n;i++)  //相当于物品
        {
            memset(num,0,sizeof(num));  //记录某个物品用了几次
            for(int j=a[i];j<=m;j++)  //相当于包
            {
                if(!vis[j]&&vis[j-a[i]]&&num[j-a[i]]<c[i])
                {
                    vis[j]=1;
                    num[j]=num[j-a[i]]+1;
                    cnt++;
                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}