C#程序计算一个数字的阶乘中的尾随零

给定一个整数 n，编写一个函数，返回 n! 中尾随零的计数。

例子 ：

Input: n = 5
Output: 1 
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24

Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....

C#

// C# program to count
// trailing 0s in n!
using System;
  
class GFG {
  
    // Function to return trailing
    // 0s in factorial of n
    static int findTrailingZeros(int n)
    {
        // Initialize result
        int count = 0;
  
        // Keep dividing n by powers
        // of 5 and update count
        for (int i = 5; n / i >= 1; i *= 5)
            count += n / i;
  
        return count;
    }
  
    // Driver Code
    public static void Main()
    {
        int n = 100;
        Console.WriteLine("Count of trailing 0s in " 
                + n + "! is " + findTrailingZeros(n));
    }
}
  
// This code is contributed by vt_m

输出：

Count of trailing 0s in 100! is 24

有关详细信息，请参阅有关在一个数字的阶乘中计算尾随零的完整文章！