HDU_2844_(多重背包)

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13453    Accepted Submission(s): 5382

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

 

一看题，以为普普通通一道模版题，结果卡时间卡到哭。。。

 

注意：当val[i]*num[i]>m时，使用完全背包更快，最开始每晚i注意到这点，通通使用二进制优化，结果超时超哭，然后发现完全背包比转化为二进制优化的01背包更快。

 

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int val[100005],num[100005];
int dp[100005],m;

int main()
{
    int n;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        memset(dp,0,sizeof(dp));
        //dp[0]=1;
        int i;
        for(i=0; i<n; i++)
            scanf("%d",&val[i]);
        for(i=0; i<n; i++)
            scanf("%d",&num[i]);
        for( i=0; i<n; i++)
        {
            int j;
            if(val[i]*num[i]>=m)
            {
                for( j=val[i];j<=m;j++)
                    dp[j]=max(dp[j],dp[j-val[i]]+val[i]);
                continue;
            }
            int tmp=1;
            while(num[i]>=tmp)
            {
                for( j=m;j>=val[i]*tmp;j--)
                    dp[j]=max(dp[j],dp[j-val[i]*tmp]+val[i]*tmp);
                num[i]-=tmp;
                tmp=tmp<<1;
            }
            for( j=m;j>=val[i]*num[i];j--)
                dp[j]=max(dp[j],dp[j-val[i]*num[i]]+val[i]*num[i]);
        }
        int cnt=0;
        for(i=1; i<=m; i++)
            if(dp[i]==i)
                cnt++;
        printf("%d
",cnt);
    }
    return 0;
}