SGU184——Patties

最新推荐文章于 2021-08-11 21:34:29 发布

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派特计划在他的生日派对上为宾客提供自制的特色卷心菜馅饼。为了准备足够的馅饼，他需要计算出利用现有的面粉、牛奶和卷心菜可以制作的最大馅饼数量。本篇介绍了一个简单的程序来解决这个问题。

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Petya is well-known with his famous cabbage patties. Petya's birthday will come very soon, and he wants to invite as many guests as possible. But the boy wants everybody to try his specialty of the house. That's why he needs to know the number of the patties he can cook using the stocked ingredients. Petya has P grams of flour, M milliliters of milk and C grams of cabbage. He has plenty of other ingredients. Petya knows that he needs K grams of flour, R milliliters of milk and V grams of cabbage to cook one patty. Please, help Petya calculate the maximum number of patties he can cook.

Input

The input file contains integer numbers P, M, C, K, R and V, separated by spaces and/or line breaks (1 <= P, M, C, K, R, V <= 10000).

Output

Output the maximum number of patties Petya can cook.

Sample test(s)

Input

3000 1000 500 30 15 60

Output

题目说是有a,b,c三个量、还有其对应的的，d，e，f三个量、每进行一场party，需消耗d e f个单位的值、问能举行几场party。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
int main()
{
    int a,b,c;
    int d,e,f;
    int m,n,k;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        scanf("%d%d%d",&d,&e,&f);
        m=(double)a/d;
        n=(double)b/e;
        k=(double)c/f;
        int w=min(m,n);
        int p=min(w,k);
        printf("%d\n",p);
    }
    return 0;
}

小水怡情~