Codeforces Round #107 (Div. 2) A（简单模拟）

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本文探讨了在冬季聚会中如何合理利用软饮、柠檬和盐等资源，通过算法实现每名参与者都能享受最多数量的特制饮品。通过输入参数包括人数、软饮瓶数、每瓶容量、柠檬数量、切片数、盐克数以及个人需求量，算法计算出每位参与者最多可以享用的饮品数量。

题目：

This winter is so cold in Nvodsk! A group of n friends decided to buy k bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has l milliliters of the drink. Also they bought c limes and cut each of them into d slices. After that they found p grams of salt.

To make a toast, each friend needs nl milliliters of the drink, a slice of lime and np grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?

Input

The first and only line contains positive integers n, k, l, c, d, p, nl, np, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.

Output

Print a single integer — the number of toasts each friend can make.

Sample test(s)

input
3 4 5 10 8 100 3 1

output
2

input
5 100 10 1 19 90 4 3

output
3

input
10 1000 1000 25 23 1 50 1

output
0

Note

A comment to the first sample:

Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is min(6, 80, 100) / 3 = 2.

题意分析：

题目大概就是n个人需要制作饮料并且尽可能的平均分配，他们有k瓶软饮料，每瓶有l毫升，c个柠檬每个能分成d片，还有p克盐。制作一瓶特制饮料需要nl毫升软饮料，一片柠檬和np克盐。不多说，水题，直接上代码。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>

using namespace std;


int main()
{
    int n,k,l,c,d,p,nl,np,ans2,ans1;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%d%d%d%d%d%d%d",&k,&l,&c,&d,&p,&nl,&np);
        ans1=min(k*l/nl/n,c*d/n);
        ans2=min(ans1,p/np/n);
        printf("%d\n",ans2);
    }
}