Given three positive integer **n**, **x**, **y**. The task is to print Greatest Common Divisor of numbers formed by n repeating x times and number formed by n repeating y times.

0 <= n, x, y <= 1000000000.

Examples:

Input : n = 123, x = 2, y = 3. Output : 123 Number formed are 123123 and 123123123. Greatest Common Divisor of 123123 and 123123123 is 123. Input : n = 4, x = 4, y = 6. Output : 44

The idea is based on Euclidean algorithm to compute GCD of two number.

Let f(n, x) be a function which gives a number n repeated x times. So, we need to find GCD(f(n, x), f(n, y)).

Let n = 123, x = 3, y = 2.

So, first number A is f(123, 3) = 123123123 and second number B is f(123, 2) = 123123. We know, GCD(A,B) = GCD(A – B, B), using this property we can substract any multiple of B, say B’ from first A as long as B’ is smaller than A.

So, A = 123123123 and B’ can be 123123000. On substracting A will became 123 and B remains same.

Therfore, A = A – B’ = f(n, x – y).

So, GCD(f(n, x), f(n, y)) = GCD(f(n, x – y), f(n, y))

We can conclude following,

GCD(f(n, x), f(n, y)) = f(n, GCD(x, y)).

Below is C++ implementation based on this approach:

// C++ program to print Greatest Common Divisor // of number formed by N repeating x times and // y times. #include<bits/stdc++.h> using namespace std; // Return the Greatest common Divisor of two numbers. int gcd(int a, int b) { if (a == 0) return b; return gcd(b%a, a); } // Prints Greatest Common Divisor of number formed // by n repeating x times and y times. void findgcd(int n, int x, int y) { // Finding GCD of x and y. int g = gcd(x,y); // Print n, g times. for (int i = 0; i < g; i++) cout << n; } // Driven Program int main() { int n = 123, x = 5, y = 2; findgcd(n, x, y); return 0; }

Output:

123

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