N (N–2)(N–4)*…中尾随零的数量。
给定一个整数 N ，任务是找出 f(N) 十进制记数法中尾随零的个数，其中 f(N) = 1 如果 N < 2 和f(N)= N * f(N–2)如果 N ≥ 2
示例:
输入:N = 12 T3】输出:1 f(12)= 12 * 10 * 8 * 6 * 4 * 2 = 46080
输入:N = 7 T3】输出: 0
方法:用十进制表示 f(N) 时尾随零的个数是 f(N) 被 2 整除的次数和 f(N) 被 5 整除的次数。有两种情况:
- 当 N 为奇数时，则 f(N) 是一些奇数的乘积，因此它不会在 2 处断开。所以答案总是 0 。
- 当 N 为偶数时，则 f(N) 可以表示为2(1 * 2 * 3 ……。 N/2) 。 f(N) 被 2 整除的次数大于被 5 整除的次数，所以只考虑被 5 整除的次数。现在，这个问题类似于计算一个数的阶乘中的尾随零。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count of
// trailing 0s in the given function
function findTrailingZeros(n)
{
// If n is odd
if (n & 1)
return 0;
// If n is even
else
{
let ans = 0;
// Find the trailing zeros
// in n/2 factorial
n = parseInt(n / 2);
while (n)
{
ans += parseInt(n / 5);
n = parseInt(n / 5);
}
// Return the required answer
return ans;
}
}
// Driver code
let n = 12;
document.write(findTrailingZeros(n));
// This code is contributed by subhammahato348
</script>
Output:
1
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