LightOJ 1001 – Opposite Task_Solution

//Tanzila Islam

//Southeast University

//mail : tanzilamohita@gmail.com



#include <iostream>

using namespace std;


int main() {

    int t;


    cin >> t;


    for (int i = 1 ; i <= t; i++) {

        int n1, n2, n3;

        cin >> n1;

        n2 = n1 / 2;

        n3 = n1 - n2;


        cout << n2 << " " << n3 << endl;

     }


     return 0;


 }