Find the last two digits of Factorial of a given Number

Given an integer N, the task is to find the last two digits of factorial of a number. 
Examples:

Input: N = 7 
Output: 40 
Explanation: 7! = 5040
Input: N = 11 
Output: 00 

Approach: We can observe that for N >= 10, the last two places of its factorial will contain 0’s only. Hence N! % 100 for any N >= 10 will always be 0. So we just calculate the factorial if N < 10 and extract the final two digits. 
Below is the implementation of above approach:

C++

// C++ implementation to
// find last two digits
// factorial of a given number

#include <bits/stdc++.h>
using namespace std;

// Function to print the
// last two digits of N!
void lastTwoDigits(long long N)
{

    // For N >= 10, N! % 100
    // will always be 0
    if (N >= 10) {
        cout << "00";
        return;
    }

    long long fac = 1;
    // Calculating N! % 100
    for (int i = 1; i <= N; i++)
        fac = (fac * i) % 100;

    cout << fac;
}

// Driver code
int main()
{
    int N = 7;
    lastTwoDigits(N);
}

Java

// Java implementation to
// find last two digits
// factorial of a given number
import java.util.*;
class GFG{

// Function to print the
// last two digits of N!
static void lastTwoDigits(double N)
{

    // For N >= 10, N! % 100
    // will always be 0
    if (N >= 10)
    {
        System.out.print("00");
        return;
    }

    double fac = 1;

    // Calculating N! % 100
    for (int i = 1; i <= N; i++)
        fac = (fac * i) % 100;

    System.out.print(fac);
}

// Driver code
public static void main(String args[])
{
    int N = 7;
    lastTwoDigits(N);
}
}

// This code is contributed by Code_Mech

Python3

# Python3 implementation to
# find last two digits
# factorial of a given number

# Function to print the
# last two digits of N!
def lastTwoDigits(N):

    # For N >= 10, N! % 100
    # will always be 0
    if (N >= 10):
        print("00", end = "")
        return

    fac = 1

    # Calculating N! % 100
    for i in range(1, N + 1):
        fac = (fac * i) % 100

    print(fac)

# Driver code
if __name__ == '__main__':
    N = 7
    lastTwoDigits(N)

# This code is contributed by Mohit Kumar

C#

// C# implementation to
// find last two digits
// factorial of a given number
using System;
class GFG{

// Function to print the
// last two digits of N!
static void lastTwoDigits(double N)
{

    // For N >= 10, N! % 100
    // will always be 0
    if (N >= 10)
    {
        Console.Write("00");
        return;
    }

    double fac = 1;

    // Calculating N! % 100
    for (int i = 1; i <= N; i++)
        fac = (fac * i) % 100;

    Console.Write(fac);
}

// Driver code
public static void Main()
{
    int N = 7;
    lastTwoDigits(N);
}
}

// This code is contributed by Nidhi_biet

Javascript

<script>

// JavaScript implementation to
// find last two digits of
// factorial of a given number

// Function to print the
// last two digits of N!
function lastTwoDigits(N)
{

    // For N >= 10, N! % 100
    // will always be 0
    if (N >= 10) {
        cout << "00";
        return;
    }

    let fac = 1;
    // Calculating N! % 100
    for (let i = 1; i <= N; i++)
        fac = (fac * i) % 100;

    document.write(fac);
}

// Driver code
    let N = 7;
    lastTwoDigits(N);


// This code is contributed by Surbhi Tyagi.

</script>

Output:

40

Time complexity: O(n) since using a for loop

Auxiliary space: O(1) because it is using constant space