include <bits/stdc++.h> using namespace std; int main() { int T; cin >> T; while (T–) { int X, Y; cin >> X >> Y;
while (X != 0) { if (X > Y) { swap(X, Y); } else { X = Y - X; Y = X; } } cout << Y << endl; } return 0;
}
can you give problem link
As the value of x and y can be upto 10^9 doing this in while loop will give tle what we are finding is the gcd of x and y you can directly use builtin function (gcd(x,y)) or { while(x%y!=0){ int z=x%y; x=y; y=z; } this will solve your problem
Test cases that gives tle are like 1000000000 1 doing this in while loop will take 10^9 operation which is greater than 1 sec