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Codeforces Round #107 (Div. 2)---A. Soft Drinking

来源:程序员人生   发布时间:2014-11-10 08:42:52 阅读次数:2563次

Soft Drinking
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 nklcdpnlnp, 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.







解题思路:贪心。不过是小学数学题。。。算出3种成份的最大值,取其中的最小值便可。






AC代码:

#include <iostream> #include <cstring> #include <cstdio> using namespace std; int main(){ // freopen("in.txt", "r", stdin); int n,k,l,c,d,p,nl,np; while(cin>>n>>k>>l>>c>>d>>p>>nl>>np){ int a = (k * l) / nl; int aa = p / np; int aaa = c * d; int ans = min(a, min(aa, aaa)) / n; cout<<ans<<endl; } }



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