Numbers whose factorials end with n zeros

Given an integer n, we need to find the number of positive integers whose factorial ends with n zeros.

Examples:

Input : n = 1
Output : 5 6 7 8 9
Explanation: Here, 5! = 120, 6! = 720,
7! = 5040, 8! = 40320 and 9! = 362880.

Input : n = 2
Output : 10 11 12 13 14 
          



Prerequisite : Trailing zeros in factorial.

Naive approach:We can just iterate through the range of integers and find the number of trailing zeros of all the numbers and print the numbers with n trailing zeros.

Efficient Approach:In this approach we use binary search. Use binary search for all the numbers in the range and get the first number with n trailing zeros. Find all the numbers with m trailing zeros after that number.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// Binary search based CPP program to find
// numbers with n trailing zeros.
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate trailing zeros
int trailingZeroes(int n)
{
    int cnt = 0;
    while (n > 0) {
        n /= 5;
        cnt += n;
    }
    return cnt;
}
  
void binarySearch(int n)
{
    int low = 0;
    int high = 1e6; // range of numbers
  
    // binary search for first number with 
    // n trailing zeros
    while (low < high) {
        int mid = (low + high) / 2;
        int count = trailingZeroes(mid);
        if (count < n)
            low = mid + 1;
        else
            high = mid;
    }
  
    // Print all numbers after low with n
    // trailing zeros.
    vector<int> result;
    while (trailingZeroes(low) == n) {
        result.push_back(low);
        low++;
    }
  
    // Print result
    for (int i = 0; i < result.size(); i++) 
        cout << result[i] << " ";
}
  
// Driver code
int main()
{
    int n = 2;
    binarySearch(n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Binary search based Java 
// program to find numbers 
// with n trailing zeros.
import java.io.*;
  
class GFG {
  
    // Function to calculate 
    // trailing zeros
    static int trailingZeroes(int n)
    {
        int cnt = 0;
        while (n > 0
        {
            n /= 5;
            cnt += n;
        }
        return cnt;
    }
  
    static void binarySearch(int n)
    {
        int low = 0;
          
        // range of numbers
        int high = 1000000;
  
        // binary search for first number 
        // with n trailing zeros
        while (low < high) {
            int mid = (low + high) / 2;
            int count = trailingZeroes(mid);
            if (count < n)
                low = mid + 1;
            else
                high = mid;
        }
  
        // Print all numbers after low 
        // with n trailing zeros.
        int result[] = new int[1000];
        int k = 0;
        while (trailingZeroes(low) == n) {
            result[k] = low;
            k++;
            low++;
        }
  
        // Print result
        for (int i = 0; i < k; i++)
            System.out.print(result[i] + " ");
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 3;
        binarySearch(n);
    }
}
  
// This code is contributed 
// by Nikita Tiwari.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Binary search based Python3 code to find
# numbers with n trailing zeros.
  
# Function to calculate trailing zeros
def trailingZeroes( n ):
    cnt = 0
    while n > 0:
        n =int(n/5)
        cnt += n
    return cnt
  
def binarySearch( n ):
    low = 0
    high = 1e6  # range of numbers
      
    # binary search for first number with
    # n trailing zeros
    while low < high:
        mid = int((low + high) / 2)
        count = trailingZeroes(mid)
        if count < n:
            low = mid + 1
        else:
            high = mid
              
    # Print all numbers after low with n
    # trailing zeros.
    result = list()
    while trailingZeroes(low) == n:
        result.append(low)
        low+=1
      
    # Print result
    for i in range(len(result)):
        print(result[i],end=" ")
  
# Driver code
n = 2
binarySearch(n)
  
# This code is contributed by "Sharad_Bhardwaj".

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// Binary search based C# 
// program to find numbers 
// with n trailing zeros.
using System;
  
class GFG {
  
    // Function to calculate 
    // trailing zeros
    static int trailingZeroes(int n)
    {
        int cnt = 0;
          
        while (n > 0) 
        {
            n /= 5;
            cnt += n;
        }
          
        return cnt;
    }
  
    static void binarySearch(int n)
    {
        int low = 0;
          
        // range of numbers
        int high = 1000000;
  
        // binary search for first number 
        // with n trailing zeros
        while (low < high) {
            int mid = (low + high) / 2;
            int count = trailingZeroes(mid);
              
            if (count < n)
                low = mid + 1;
            else
                high = mid;
        }
  
        // Print all numbers after low 
        // with n trailing zeros.
        int []result = new int[1000];
        int k = 0;
        while (trailingZeroes(low) == n) {
            result[k] = low;
            k++;
            low++;
        }
  
        // Print result
        for (int i = 0; i < k; i++)
            Console.Write(result[i] + " ");
    }
  
    // Driver code
    public static void Main()
    {
        int n = 2;
          
        binarySearch(n);
    }
}
  
// This code is contributed by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php 
// Binary search based PHP program to 
// find numbers with n trailing zeros.
  
// Function to calculate trailing zeros
function trailingZeroes($n)
{
    $cnt = 0;
    while ($n > 0) 
    {
        $n = intval($n / 5);
        $cnt += $n;
    }
    return $cnt;
}
  
function binarySearch($n)
{
    $low = 0;
    $high = 1e6; // range of numbers
  
    // binary search for first number 
    // with n trailing zeros
    while ($low < $high)
    {
        $mid = intval(($low + $high) / 2);
        $count = trailingZeroes($mid);
        if ($count < $n)
            $low = $mid + 1;
        else
            $high = $mid;
    }
  
    // Print all numbers after low with n
    // trailing zeros.
    $result = array();
    while (trailingZeroes($low) == $n)
    {
        array_push($result, $low);
        $low++;
    }
  
    // Print result
    for ($i = 0; 
         $i < sizeof($result); $i++) 
        echo $result[$i] . " ";
}
  
// Driver code
$n = 2;
binarySearch($n);
  
// This code is contributed by Ita_c
?>

chevron_right



Output:

10 11 12 13 14 


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : chitranayal