N 的 16 进制表示中尾随零的个数！

原文:https://www . geeksforgeeks . org/n 的 16 进制表示的尾随零数/

给定一个整数 N ，任务是在 N 的阶乘的 16 进制表示中找到尾随零的数量。 例:

输入:N = 6 T3】输出: 1 6！= 720(基数 10) = 2D0(基数 16) 输入: N = 100 输出: 24

进场:

  • 尾随零的数量将是基数 10 中 N 的阶乘 16 的最高幂。
  • 我们知道16 = 24T3。所以， 16 的最高幂等于 N 除以 4 的阶乘中的最高幂 2 。
  • 计算 N 中 2 的最高幂！，我们可以用勒让德公式。

以下是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to return the count of trailing zeroes
ll getTrailingZeroes(ll n)
{
    ll count = 0;
    ll val, powerTwo = 2;

    // Implementation of the Legendre's formula
    do {
        val = n / powerTwo;
        count += val;
        powerTwo *= 2;
    } while (val != 0);

    // Count has the highest power of 2
    // that divides n! in base 10
    return (count / 4);
}

// Driver code
int main()
{
    int n = 6;
    cout << getTrailingZeroes(n);
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java implementation of the approach
class GfG
{

// Function to return the count of trailing zeroes
static long getTrailingZeroes(long n)
{
    long count = 0;
    long val, powerTwo = 2;

    // Implementation of the Legendre's formula
    do
    {
        val = n / powerTwo;
        count += val;
        powerTwo *= 2;
    } while (val != 0);

    // Count has the highest power of 2
    // that divides n! in base 10
    return (count / 4);
}

// Driver code
public static void main(String[] args)
{
    int n = 6;
    System.out.println(getTrailingZeroes(n));
}
}

// This code is contributed by
// Prerna Saini.

Python 3

# Python3 implementation of the approach

# Function to return the count of
# trailing zeroes
def getTrailingZeroes(n):

    count = 0
    val, powerTwo = 1, 2

    # Implementation of the Legendre's
    # formula
    while (val != 0):
        val = n //powerTwo
        count += val
        powerTwo *= 2

    # Count has the highest power of 2
    # that divides n! in base 10
    return (count // 4)

# Driver code
n = 6
print(getTrailingZeroes(n))

# This code is contributed
# by Mohit Kumar

C

// C# implementation of the approach
using System;
class GFG
{

// Function to return the count of
// trailing zeroes
static long getTrailingZeroes(long n)
{
    long count = 0;
    long val, powerTwo = 2;

    // Implementation of the
    // Legendre's formula
    do
    {
        val = n / powerTwo;
        count += val;
        powerTwo *= 2;
    } while (val != 0);

    // Count has the highest power of 2
    // that divides n! in base 10
    return (count / 4);
}

// Driver code
public static void Main()
{
    int n = 6;
    Console.Write(getTrailingZeroes(n));
}
}

// This code is contributed by
// Akanksha Rai

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?php
// PHP implementation of the approach

// Function to return the count of
// trailing zeroes
function getTrailingZeroes($n)
{
    $count = 0;
    $val; $powerTwo = 2;

    // Implementation of the Legendre's formula
    do
    {
        $val = (int)($n / $powerTwo);
        $count += $val;
        $powerTwo *= 2;
    } while ($val != 0);

    // Count has the highest power of 2
    // that divides n! in base 10
    return ($count / 4);
}

// Driver code
$n = 6;
echo(getTrailingZeroes($n));

// This code is contributed by
// Code_Mech.

?>

java 描述语言

<script>

// JavaScript implementation of the approach

// Function to return the count of trailing zeroes
function getTrailingZeroes(n)
{
    let count = 0;
    let val, powerTwo = 2;

    // Implementation of the Legendre's formula
    do {
        val = Math.floor(n / powerTwo);
        count += val;
        powerTwo *= 2;
    } while (val != 0);

    // Count has the highest power of 2
    // that divides n! in base 10
    return (Math.floor(count / 4));
}

// Driver code

    let n = 6;
    document.write(getTrailingZeroes(n));

// This code is contributed by Surbhi Tyagi

</script>

Output: 

1

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