hdu2844Coins【多重背包】

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
   
   
   
   

 

比模板多了的唯一一点是查多少数字出现过==很容易想到枚举一遍dp[i]==i的个数

1A呦 真是幸运的一个晚上，会不会好运从此开始了呢戒骄戒躁 一切都会好起来的==

/*********
hud2844
2015.11.2
390MS 1956K 1298 B
*********/
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[100007],n,m,a[105],c[105],count;
void zeropack(int cost,int value)
{
    for(int i=m;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+value);
   // cout<<"01"<<endl;
}
void completepack(int cost,int value)
{
    for(int i=cost;i<=m;i++)
        dp[i]=max(dp[i],dp[i-cost]+value);
    //cout<<"complete"<<endl;
}
void multipack(int cost,int value,int num)
{
    if(num*cost>=m)
    {
        completepack(cost,value);
        return;
    }
    int k=1;
    while(k<num)
    {
        zeropack(k*cost,k*value);
        num-=k;
        k*=2;
    }
    zeropack(num*cost,num*value);
}
int main()
{
   // freopen("cin.txt","r",stdin);
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0) break;
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++) scanf("%d",&c[i]);
        memset(dp,0,sizeof(dp));
        count=0;
        for(int i=1;i<=n;i++)
        {
            multipack(a[i],a[i],c[i]);
        }

        for(int i=1;i<=m;i++)
        {
            if(dp[i]==i) count++;
            //printf("dp[]=%d\n",dp[i]);
        }
        printf("%d\n",count);
    }
    return 0;
}