Decreasing Srrnmieeda CodeChef October CookOff Editorial and Python solution

Decreasing Srrnmieeda Editorial and Solutions

PROBLEM STATEMENT

YOU CAN FIND THE COMPLETE PROBLEM STATEMENT HERE :PROBLEM STATEMENT

You are given two integers L$L$ and R$R$. Find the smallest non-negative integer N$N$ such that

N%L>N%(L+1)>…>N%(R−1)>N%R.

Here, %$\mathrm{\%}$ is the modulo operator, so A%B$A\mathrm{\%}B$ is the remainder of A$A$ after division by B$B$. For example, 11%3=2$11\mathrm{\%}3=2$.

UNDERSTANDING THE PROBLEM:

So We are given two integers as the input L and R respectively, now our job is to find the minimum possible non-negative integer such that :

N%L>N%(L+1)>…>N%(R−1)>N%R.

So If We take any number we will notice that the largest modulo is given by number+(number-1) because after this we will get 2*number which will give the modulo as 0. If we now increase the number we will notice that the remainder starts repeating themselves.

So by this logic the value of r should not be greater than the number which gives the largest modulo which is 1 less than twice the number so:

if r is less than twice l then we will get the answer or else its impossible to get such an n as if its greater than n it would be increasing and then decrease after reaching the highest remainder.

Now in the same way we will get the minimum n only if the remainder is 0. because if n gives remainder 1 with the lower number it means that there is another value of n and as the function asks us to find the minimum value of n we can safely say that the minimum value of n will be the value of r

PYTHON 3 CODE:

Now that we know the logic behind the code lets implement it 

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# cook your dish testcases = int(input()) for i in range(testcases): a = input() num1,num2 = a.split(" ") num1,num2 = int(num1),int(num2) if num2<2*num1: print(num2) else: print(-1)