HDU Flowers 完全背包

Flowers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2709    Accepted Submission(s): 1811

Problem Description

As you know, Gardon trid hard for his love-letter, and now he's spending too much time on choosing flowers for Angel.
When Gardon entered the flower shop, he was frightened and dazed by thousands kinds of flowers. 
"How can I choose!" Gardon shouted out.
Finally, Gardon-- a no-EQ man decided to buy flowers as many as possible.
Can you compute how many flowers Gardon can buy at most?

 

Input

Input have serveral test cases. Each case has two lines.
The first line contains two integers: N and M. M means how much money Gardon have.
N integers following, means the prices of differnt flowers.

 

Output

For each case, print how many flowers Gardon can buy at most.
You may firmly assume the number of each kind of flower is enough.

 

Sample Input

2 5 2 3

 

Sample Output

2

Hint

Hint

Gardon can buy 5=2+3,at most 2 flower, but he cannot buy 3 flower with 5 yuan.

 

Author

DYGG

 

Source

HDU “Valentines Day” Open Programming Contest 2007-02-14

 

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linle

 

裸完全背包，拥有的钱为背包容量M，每一枝花的花费为1，价值价值为w[i].

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int a[1000000];
int dp[1000000];
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n;i++)
            for(int j=a[i];j<=m;j++)
                dp[j]=max(dp[j],dp[j-a[i]]+1);
        printf("%d
",dp[m]);
    }
    return 0;
}