Soft Drinking(水)

最新推荐文章于 2025-08-13 10:25:05 发布

转载最新推荐文章于 2025-08-13 10:25:05 发布 · 163 阅读

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#数据结构与算法

一群朋友在寒冷的冬天购买软饮和柠檬汁来取暖，通过计算每名朋友所需的软饮量、柠檬片数及盐量，以确定他们能制作多少杯酒。输入包括人数、软饮瓶数、每瓶容量、柠檬数量、切片数、盐克数、所需软饮量和盐量，输出为每名朋友能制作的酒杯数。

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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.

Examples

input

3 4 5 10 8 100 3 1

output

input

5 100 10 1 19 90 4 3

output

input

10 1000 1000 25 23 1 50 1

output

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 = 80 toasts 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.

题解：水题：配饮料的，问每个人能配几瓶；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")

int main(){
    int n,k,l,c,d,p,nl,np;
    while(~scanf("%d%d%d%d%d%d%d%d",&n,&k,&l,&c,&d,&p,&nl,&np)){
        printf("%d\n",min(min(k*l/nl,c*d),p/np)/n);
    }
    return 0;
}