HDU1587 Flowers【水题】

Flowers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4189    Accepted Submission(s): 2753

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

问题链接：HDU1587 Flowers。

题意简述：输入n和m，然后输入n种花的价格。问n种花，金额m，最多可以买多少花。

问题分析：找出最小（最便宜）价格，用金额除以最小价格即是答案。

程序说明：（略）

题记：（略）

AC的C语言程序如下：

/* HDU1587 Flowers */

#include <stdio.h>
#include <limits.h>

int main(void)
{
    int n, m, a, min, i;

    while(scanf("%d%d",&n, &m) != EOF) {
        min = INT_MAX;
        for(i=1; i<=n; i++) {
            scanf("%d", &a);
            if(a < min)
                min = a;
        }

        printf("%d
", m / min);
    }

    return 0;
}