hdu2844 Coins (多重背包+二进制优化)

最新推荐文章于 2022-09-09 22:07:37 发布

Mr_Treeeee

最新推荐文章于 2022-09-09 22:07:37 发布

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分类专栏： HDU 背包

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本文介绍了一道关于硬币组合的编程题目，使用多重背包二进制优化算法解决硬币支付问题，计算出所有可能支付的价格数目。

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Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14518 Accepted Submission(s): 5757

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

题意：

硬币能组成不大于price of the watch的价钱的数量。

point：

用多重背包二进制优化。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int dp[100200];
int val[102];
int num[102];
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        if(n==0&&m==0) break;
        memset(dp,0,sizeof dp);
        for(int i=1;i<=n;i++) scanf("%d",&val[i]);
        for(int i=1;i<=n;i++) scanf("%d",&num[i]);
        for(int i=1;i<=n;i++)
        {
            int now=num[i];
            int k=1;
            while(k<now)
            {
                now-=k;
                for(int j=m;j>=k*val[i];j--)
                {
                    if(dp[j]<j)
                        dp[j]=max(dp[j],dp[j-k*val[i]]+k*val[i]);
                }
                k*=2;
            }
            for(int j=m;j>=now*val[i];j--)
            {
                if(dp[j]<j)
                    dp[j]=max(dp[j],dp[j-now*val[i]]+now*val[i]);
            }

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

    return 0;
}