Editorial for XORNEY - PELT2019

cakewalk
pelt2019
tds115
xorney

#1

Question Code: XORNEY

Prerequisites: Implementation

Problem Setter: tds115

Difficulty: Cakewalk

Explanation:
Given two integer l and r, you have to find if xor from l to r is even or odd. Basically, you have
to find the number of odd integer between l and r. Since xor of odd 1’s is 1 and xor of even
1’s is 0. You can make 2 cases:-
(Let count be the number of odd’s)
If l is even and r is even-> count=(r-l)/2
Else -> count=(r-l)/2+1
If count is even then ans is even
Else odd.

Complexity-O(1) for each query

Author’s solution: https://ideone.com/GUjqC5