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
 
Recommend
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].
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<stdlib.h>
 4 #include<algorithm>
 5 using namespace std;
 6 int a[1000000];
 7 int dp[1000000];
 8 int main()
 9 {
10     int n,m;
11     while(scanf("%d %d",&n,&m)!=EOF)
12     {
13         memset(dp,0,sizeof(dp));
14         for(int i=0;i<n;i++)
15             scanf("%d",&a[i]);
16         for(int i=0;i<n;i++)
17             for(int j=a[i];j<=m;j++)
18                 dp[j]=max(dp[j],dp[j-a[i]]+1);
19         printf("%d\n",dp[m]);
20     }
21     return 0;
22 }
View Code

 

posted @ 2014-08-03 21:33  Cliff Chen  阅读(225)  评论(0编辑  收藏  举报