给定三个正整数n ， x ， y 。任务是打印n次重复x次形成的数字和n次重复y次形成的数字的最大公约数。
0 <= n，x，y <= 1000000000。
例子 ：
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这个想法是基于欧几里德算法来计算两个数的GCD。
令f(n，x)是一个将n重复x次的函数。因此，我们需要找到GCD(f(n，x)，f(n，y))。
令n = 123，x = 3，y = 2。
因此，第一个数字A为f(123，3)= 123123123，第二个数字B为f(123，2)=123123。我们知道，使用此属性，GCD(A，B)= GCD(A – B，B)只要B’小于A，我们就可以从第一个A中减去B的任何倍数，例如B’。
因此，A = 123123123，而B’可以为123123000。减去A将变为123，而B保持不变。
因此，A = A – B’= f(n，x – y)。
因此，GCD(f(n，x)，f(n，y))= GCD(f(n，x – y)，f(n，y))
我们可以得出以下结论：
GCD(f(n, x), f(n, y)) = f(n, GCD(x, y)). 以下是基于此方法的实现：
CPP
// C++ program to print Greatest Common Divisor
// of number formed by N repeating x times and
// y times.
#include
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;
} Java
// Java program to print Greatest Common Divisor
// of number formed by N repeating x times and
// y times
class GFG {
// Return the Greatest common Divisor
// of two numbers.
static 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.
static 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++)
System.out.print(n);
}
// Driver code
public static void main(String[] args) {
int n = 123, x = 5, y = 2;
findgcd(n, x, y);
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to print Greatest
# Common Divisor of number formed
# by N repeating x times and y times
# Return the Greatest common Divisor
# of two numbers.
def gcd(a, 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.
def findgcd(n, x, y):
# Finding GCD of x and y.
g = gcd(x, y)
# Print n, g times.
for i in range(g):
print(n)
# Driver code
n = 123
x = 5
y = 2
findgcd(n, x, y)
# This code is contributed by Anant Agarwal.C#
// C# program to print Greatest Common
// Divisor of number formed by N
// repeating x times and y times
using System;
class GFG {
// Return the Greatest common
// Divisor of two numbers.
static 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.
static 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++)
Console.Write(n);
}
// Driver code
public static void Main() {
int n = 123, x = 5, y = 2;
findgcd(n, x, y);
}
}
// This code is contributed by
// nitin mittal.PHP
Javascript
输出 ：
123