Leyni, LOLI and Toasts

Source: Internet
Author: User

Leyni, LOLI and Toasts
Time limit:1000 MS Memory limit:65536 K
Total submit:47(Notoginseng users) Total accepted:37(Notoginseng users) Rating: Special Judge: No

Leyni likes to play with Lolis, but this winter are so cold in harbin!

A Group of N Lolis decided to buy b bottles of soft drink to make themselves warmer. Each bottle has D milliliters of the drink. Also They bought l Limes and cut each of the them into k slices. After that they found p grams of salt.

To make a toast, each LOLI needs x milliliters of the drink, a slice of lime and y grams of salt. Have the average materials, the lolis want to make toasts as many as they can. How many toasts can each LOLI makes?


There is multiple test cases. The first line of input was an integer T indicating the number of test cases. Then T test Cases follow.

For each test case:

Line 1. This line contains eight positive integers, n, b, D, l, K, p, x< /c8>, y (1≤ N, b, D, L, K, p, x, y ≤1000). The numbers is separated by exactly one space.


For each test case:

Line 1. Output the number of toasts each LOLI can make.

Sample Input


3 4 5 10 8 100 3 1

Sample Output



In the sample, overall the Lolis has 4 * 5 = milliliters of the drink, it is enough to make 20/3 = 6 toasts. The limes was enough for 8 = toasts and the salt was enough for 100/1 = toasts. However, there is 3 Lolis in the group and so the answer is 6/3 = 2.

Harbin Tech 2012 Spring School race-Live race
Zidaratu @hrbust
#include <cstdio>#include<cmath>#include<queue>#include<iostream>using namespacestd;intMinintXinty) {    returnX<y?x:y;}intMain () {intN, B, D, L, K, p, X, y; intT;  while(~SCANF ("%d",&T)) { while(t--) {scanf ("%d%d%d%d%d%d%d%d",&n,&b,&d,&l,&k,&p,&x,&y); /*if (n>b) {printf ("0\n"); }*/            //Else//    {                intNum1=b*d/(nx); intnum2=l*k/N; intnum3=p/(y*N); printf ("%d\n", Min (num1,min (NUM2,NUM3))); //    }        }    }    return 0;}

Leyni, LOLI and Toasts

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