hdu 2844 Coins

Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5407 Accepted Submission(s): 2230


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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//593MS    640K    1071 B    C++
//多重背包小优化算法
//装换成01+完全，时间复杂度降低才能过 
#include<stdio.h>
#include<string.h>
int dp[100005],c[105],v[105];
int n,m;
int Max(int a,int b)
{
    return a>b?a:b;
}
void _01pack(int cost,int weight)
{
    for(int i=m;i>=weight;i--)
        dp[i]=Max(dp[i],dp[i-weight]+cost);
} 
void compack(int cost,int weight)
{
    for(int i=weight;i<=m;i++)
        dp[i]=Max(dp[i],dp[i-weight]+cost); 
}
void mulpack(int cost,int weight,int count)
{
    if(count*cost>=m){
        compack(cost,weight);return;
    }
    int k=1;
    while(k<count){
        _01pack(k*cost,k*weight);
        count-=k;
        k*=2;
    }
    _01pack(count*cost,count*weight);
} 
int main(void)
{
    while(scanf("%d%d",&n,&m),n+m)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
            scanf("%d",&v[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&c[i]);
        for(int i=0;i<n;i++)
            mulpack(v[i],v[i],c[i]);
        int cnt=0;
        for(int i=1;i<=m;i++)
            if(dp[i]==i) cnt++;
        printf("%d\n",cnt);       
    }
    return 0;
}