给定数字n 。任务是找到其阶乘包含至少n个尾随零的最小数字。
例子 ：
Input : n = 1
Output : 5
1!, 2!, 3!, 4! does not contain trailing zero.
5! = 120, which contains one trailing zero.
Input : n = 6
Output : 25在关于数的阶乘计数尾随零的文章中，我们讨论了零的个数等于x！的质数中的5的个数。我们已经讨论了以下公式来计算5的数量。
Trailing 0s in x! = Count of 5s in prime factors of x!
= floor(x/5) + floor(x/25) + floor(x/125) + ....让我们举几个例子来观察模式
5! has 1 trailing zeroes
[All numbers from 6 to 9
have 1 trailing zero]
10! has 2 trailing zeroes
[All numbers from 11 to 14
have 2 trailing zeroes]
15! to 19! have 3 trailing zeroes
20! to 24! have 4 trailing zeroes
25! to 29! have 6 trailing zeroes我们可以注意到，阶乘包含n个尾随零的最小值为5 * n。
因此，要查找其阶乘包含n个尾随零的最小值，请在0到5 * n的范围内使用二进制搜索。并且，找到其阶乘包含n个尾随零的最小数字。
C++
// C++ program tofind smallest number whose
// factorial contains at least n trailing
// zeroes.
#include
using namespace std;
// Return true if number's factorial contains
// at least n trailing zero else false.
bool check(int p, int n)
{
int temp = p, count = 0, f = 5;
while (f <= temp)
{
count += temp/f;
f = f*5;
}
return (count >= n);
}
// Return smallest number whose factorial
// contains at least n trailing zeroes
int findNum(int n)
{
// If n equal to 1, return 5.
// since 5! = 120.
if (n==1)
return 5;
// Initalising low and high for binary
// search.
int low = 0;
int high = 5*n;
// Binary Search.
while (low > 1;
// Checking if mid's factorial contains
// n trailing zeroes.
if (check(mid, n))
high = mid;
else
low = mid+1;
}
return low;
}
// driver code
int main()
{
int n = 6;
cout << findNum(n) << endl;
return 0;
} Java
// Java program tofind smallest number whose
// factorial contains at least n trailing
// zeroes.
class GFG
{
// Return true if number's factorial contains
// at least n trailing zero else false.
static boolean check(int p, int n)
{
int temp = p, count = 0, f = 5;
while (f <= temp)
{
count += temp / f;
f = f * 5;
}
return (count >= n);
}
// Return smallest number whose factorial
// contains at least n trailing zeroes
static int findNum(int n)
{
// If n equal to 1, return 5.
// since 5! = 120.
if (n==1)
return 5;
// Initalising low and high for binary
// search.
int low = 0;
int high = 5 * n;
// Binary Search.
while (low < high)
{
int mid = (low + high) >> 1;
// Checking if mid's factorial
// contains n trailing zeroes.
if (check(mid, n))
high = mid;
else
low = mid + 1;
}
return low;
}
// Driver code
public static void main (String[] args)
{
int n = 6;
System.out.println(findNum(n));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program tofind smallest
# number whose
# factorial contains at least
# n trailing zeroes
# Return true if number's factorial contains
# at least n trailing zero else false.
def check(p,n):
temp = p
count = 0
f = 5
while (f <= temp):
count += temp/f
f = f*5
return (count >= n)
# Return smallest number whose factorial
# contains at least n trailing zeroes
def findNum(n):
# If n equal to 1, return 5.
# since 5! = 120.
if (n==1):
return 5
# Initalizing low and high for binary
# search.
low = 0
high = 5*n
# Binary Search.
while (low > 1
# Checking if mid's factorial contains
# n trailing zeroes.
if (check(mid, n)):
high = mid
else:
low = mid+1
return low
# driver code
n = 6
print(findNum(n))
# This code is contributed
# by Anant Agarwal. C#
// C# program tofind smallest number whose
// factorial contains at least n trailing
// zeroes.
using System;
class GFG
{
// Return true if number's factorial contains
// at least n trailing zero else false.
static bool check(int p, int n)
{
int temp = p, count = 0, f = 5;
while (f <= temp)
{
count += temp / f;
f = f * 5;
}
return (count >= n);
}
// Return smallest number whose factorial
// contains at least n trailing zeroes
static int findNum(int n)
{
// If n equal to 1, return 5.
// since 5! = 120.
if (n == 1)
return 5;
// Initalising low and high for binary
// search.
int low = 0;
int high = 5 * n;
// Binary Search.
while (low < high)
{
int mid = (low + high) >> 1;
// Checking if mid's factorial
// contains n trailing zeroes.
if (check(mid, n))
high = mid;
else
low = mid + 1;
}
return low;
}
// Driver code
public static void Main ()
{
int n = 6;
Console.WriteLine(findNum(n));
}
}
// This code is contributed by vt_m.PHP
= $n);
}
// Return smallest number
// whose factorial contains
// at least n trailing zeroes
function findNum($n)
{
// If n equal to 1, return 5.
// since 5! = 120.
if ($n == 1)
return 5;
// Initalising low and high
// for binary search.
$low = 0;
$high = 5 * $n;
// Binary Search.
while ($low < $high)
{
$mid = ($low + $high) >> 1;
// Checking if mid's factorial
// contains n trailing zeroes.
if (check($mid, $n))
$high = $mid;
else
$low = $mid + 1;
}
return $low;
}
// Driver Code
$n = 6;
echo(findNum($n));
// This code is contributed by Ajit.
?>Javascript
输出 ：
25