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题目：

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

Recommend

gaojie

背包模板果然好用。。。

是一个多重背包转换成二进制背包的典型例子。。。

就是相当于1——m的背包能否装满。。。。即价值为i。。。

代码如下：

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=100000+10;
int dp[maxn],v[105],Count[105],m;
void zeroonepack(int cost,int value)
{
    for(int i=m;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+value);
}

void completepack(int cost,int value)
{
    for(int i=cost;i<=m;i++)
        dp[i]=max(dp[i],dp[i-cost]+value);
}

void multiplepack(int cost,int value,int amount)
{
    int k=1;
    if(cost*amount>=m)
        completepack(cost,value);
    else
    {
        while(k<amount)
        {
            zeroonepack(k*cost,k*value);
            amount-=k;
            k*=2;
        }
        zeroonepack(amount*cost,amount*value);
    }
}

int main()
{
    int i,j,n,min_count;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        min_count=0;
        if(n==0&&m==0) return 0;
        for(i=1;i<=n;i++)
            scanf("%d",&v[i]);
        for(i=1;i<=n;i++)
            scanf("%d",&Count[i]);
        for(i=1;i<=n;i++)
        {
            multiplepack(v[i],v[i],Count[i]);
        }
        for(i=1;i<=m;i++)
        {
            if(dp[i]==i)
                min_count++;
        }
        printf("%d\n",min_count);

    }
    return 0;

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