Interviews#12339

Interviews (Editorial)

Problem Author:	Hasan Saleh (MathModel)
Prepared by:	Hasan Saleh (MathModel)
Edited by:	Hasan Saleh (MathModel)

There are $x$ testers, each of them can do $y$ interviews per day, and there are $p$ participants. Each interview requires $z$ testers and 1 participant to be present. Our goal is to minimize the number of days, so we should maximize performances as follows:-

$x$ testers $\to$ $y$ interviews= $x y$ interviews in total.

Maximum performance per day is $⌊ \frac{x y}{z} ⌋$ \lfloor \frac{xy}{z} \rfloor interviews because we want to calculate the number of distinct participants tested. We divide by $z$ and perform floor to get rid of the decimal part.

Finally, we divide the number of participants by the maximum performance per day to get the minimum number of days , and we perform $ceil$ if there is a decimal part to do it the next day:

min (days) = ⌈ \frac{p}{⌊ \frac{x y}{z} ⌋} ⌉

\min(\text{days})=\left \lceil \frac{p}{\left\lfloor \frac{xy}{z} \right\rfloor} \right\rceil

If $x < z$ x<z then it's impossible.

from math import *
for _ in range(int(input())):
    p,x,y,z=map(int,input().split())
    if(x<z):print(-1)
    else:
        print(ceil(p/((x*y)//z)))