N *(N – 2)*(N – 4)*…中的尾随零数。

给定整数N ，任务是找到f(N)的十进制表示形式的尾随零个数，如果N <2 ，则f(N)= 1，如果f(N)= N * f(N – 2)，则f(N)= 1 N≥2

例子：

Input: N = 12
Output: 1
f(12) = 12 * 10 * 8 * 6 * 4 * 2 = 46080

Input: N = 7
Output: 0

方法：用十进制表示法表示f(N)时的尾随零数是f(N)被2整除的次数和f(N)被5可整除的次数。有两种情况：

1. 当N是奇数时，则f(N)是一些奇数的乘积，因此它不会在2处中断。因此，答案始终为0 。
2. 当N为偶数时， f(N)可以表示为2(1 * 2 * 3 *…。* N / 2) 。 f(N)可被2整除的次数大于可被5整除的次数，因此仅考虑可被5整除的次数。现在，此问题类似于在数字阶乘中计数尾随零。

下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of
// trailing 0s in the given function
int findTrailingZeros(int n)
{
    // If n is odd
    if (n & 1)
        return 0;
  
    // If n is even
    else {
        int ans = 0;
  
        // Find the trailing zeros
        // in n/2 factorial
        n /= 2;
        while (n) {
            ans += n / 5;
            n /= 5;
        }
  
        // Return the required answer
        return ans;
    }
}
  
// Driver code
int main()
{
    int n = 12;
  
    cout << findTrailingZeros(n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
    // Function to return the count of 
    // trailing 0s in the given function 
    static int findTrailingZeros(int n) 
    { 
        // If n is odd 
        if ((n & 1) == 1) 
            return 0; 
      
        // If n is even 
        else 
        { 
            int ans = 0; 
      
            // Find the trailing zeros 
            // in n/2 factorial 
            n /= 2; 
            while (n != 0)
            { 
                ans += n / 5; 
                n /= 5; 
            } 
      
            // Return the required answer 
            return ans; 
        } 
    } 
      
    // Driver code 
    public static void main (String[] args)
    { 
        int n = 12; 
      
        System.out.println(findTrailingZeros(n)); 
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
  
# Function to return the count of
# trailing 0s in the given function
def findTrailingZeros(n):
      
    # If n is odd
    if (n & 1):
        return 0
  
    # If n is even
    else:
        ans = 0
  
        # Find the trailing zeros
        # in n/2 factorial
        n //= 2
        while (n):
            ans += n // 5
            n //= 5
  
        # Return the required answer
        return ans
  
# Driver code
  
n = 12
  
print(findTrailingZeros(n))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
  
class GFG 
{
      
    // Function to return the count of 
    // trailing 0s in the given function 
    static int findTrailingZeros(int n) 
    { 
        // If n is odd 
        if ((n & 1) == 1) 
            return 0; 
      
        // If n is even 
        else
        { 
            int ans = 0; 
      
            // Find the trailing zeros 
            // in n/2 factorial 
            n /= 2; 
            while (n != 0)
            { 
                ans += n / 5; 
                n /= 5; 
            } 
      
            // Return the required answer 
            return ans; 
        } 
    } 
      
    // Driver code 
    public static void Main(String[] args)
    { 
        int n = 12; 
      
        Console.WriteLine(findTrailingZeros(n)); 
    } 
}
  
// This code is contributed by 29AjayKumar

输出：

1