If the input is P then the output will be 1 if P can be expressed as 1+(n(n-1)/2)

=> 1+(n(n-1)/2) = p
=> 2+n(n-1) = 2p
=> n(n-1) = 2p-2

To find such n I used floor and ceil of sqrt(2p-2)

What's wrong with my logic?

#include <bits/stdc++.h>

using namespace std;

int main(){
    double n, k, a, b, root;
    vector<int> v;
    cin >> k;
    while(k--){
        cin >> n;
        root = sqrt((2*n) - 2);
        a = ceil(root);
        b = floor(root);
        if(a == root && b == root){
            v.push_back(0);
        }else{
            if(a * b == (2*n) - 2) v.push_back(1);
            else v.push_back(0);
        }
    }
    for (int i = 0; i < v.size(); ++i){
        if(i == v.size() - 1) cout << v[i];
        else cout << v[i] << " ";
    }
    return 0;
}

Edited by author 30.08.2016 20:42