给定范围内的元素的GCD

给定两个数字n和m。找出最大的整数a(gcd)，以使所有整数n，n + 1，n + 2，…，m可被a整除。

例子：

Input : n = 1, m = 2
Output: 1
Explanation:
Here, series become 1, 2. So, the 
greatest no which divides both of 
them is 1.

Input : n = 475, m = 475
Output : 475
Explanation:
Here, series has only one term 475.
So, greatest no which divides 475 is 475.

在这里，我们只需要检查两种情况：

1. 如果a = b：该段由单个数字组成，则答案为a。
2. 如果a 下面是上述方法的代码。

   C++

   // GCD of given range
   #include 
   using namespace std;
     
   int rangeGCD(int n, int m)
   {
       return (n == m)? n : 1;
   }
     
   int main()
   {
       int n = 475;
       int m = 475;
       cout << rangeGCD(n, m);
       return 0;
   }

   Java

   // GCD of given range
     
   import java.io.*;
     
   class GFG {
     
       static int rangeGCD(int n, int m)
       {
           return (n == m) ? n : 1;
       }
     
       public static void main(String[] args)
       {
           int n = 475;
           int m = 475;
             
           System.out.println(rangeGCD(n, m));
       }
   }
     
   // This code is contributed by Ajit.

   Python3

   # GCD of given range
     
   def rangeGCD(n, m):
       return n if(n == m) else 1
         
   # Driver code
   n, m = 475, 475
   print(rangeGCD(n, m))
     
   # This code is contributed by Anant Agarwal.

   C#

   // GCD of given range
   using System;
      
   class GFG {
      
       static int rangeGCD(int n, int m)
       {
           return (n == m) ? n : 1;
       }
      
       public static void Main()
       {
           int n = 475;
           int m = 475;
              
           Console.WriteLine(rangeGCD(n, m));
       }
   }
      
   // This code is contributed by Anant Agarwal.

   输出：

   475