HackerEarth Even Odd Queries problem solution YASH PAL, 31 July 202413 February 2026 In this HackerEarth Even Odd Queries problem solution You are given an array Arr of size N, containing integers. You have to answer Q queries where each query is of the form : K L R :- If K = 0, then you have to find the probability of choosing an even number from the segment [L,R] (both Inclusive) in the array Arr. K L R :- If K = 1, then you have to find the probability of choosing an odd number from the segment [L,R] (both Inclusive) in the array Arr. For each query print two integers p and q which represent the probability p/q. Both p and q are reduced to the minimal form. If p is 0 or p is equal to q print p/q alone. HackerEarth Even Odd Queries problem solution.#include <bits/stdc++.h>#define sflld(n) scanf("%lld",&n)#define sfulld(n) scanf("%llu",&n)#define sfd(n) scanf("%d",&n)#define sfld(n) scanf("%ld",&n)#define sfs(n) scanf("%s",&n)#define ll long long#define s(t) int t; while(t--)#define ull unsigned long long int#define pflld(n) printf("%lldn",n)#define pfd(n) printf("%dn",n)#define pfld(n) printf("%ldn",n)#define lt 2*idx#define rt 2*idx+1#define f(i,k,n) for(i=k;i<n;i++)#define MAXN 100005using namespace std;int ct[MAXN][2];int main(){ int t; sfd(t); while(t--) { int n,q,i; sfd(n); sfd(q); f(i,1,n+1) { int x; sfd(x); ct[i][x%2]=ct[i-1][x%2]+1; ct[i][1-x%2]=ct[i-1][1-x%2]; } while(q--) { int l,r,k; sfd(k); sfd(l); sfd(r); int q=r-l+1; int p=ct[r][k]-ct[l-1][k]; if(p==0||p==q) cout<<p/q<<endl; else { int g=__gcd(p,q); p=p/g; q=q/g; cout<<p<<" "<<q<<endl; } } } return 0;} coding problems solutions HackerEarth HackerEarth