Extra long Factorials


Problem Statement :


The factorial of the integer n, written n!, is defined as:
         n! = n * (n-1) * (n-2) * ........ * 3 * 2 * 1
Calculate and print the factorial of a given integer.

For example, if n = 30, we calculate 30 * 29 * 28 * ...... * 3 * 2 * 1 and get 265252859812191058636308480000000.


Function Description

Complete the extraLongFactorials function in the editor below. It should print the result and return.
extraLongFactorials has the following parameter(s):
n: an integer


Note: Factorials of  can't be stored even in a  long long variable. Big integers must be used for such calculations. Languages like Java, Python, Ruby etc. can handle big integers, but we need to write additional code in C/C++ to handle huge values.

We recommend solving this challenge using BigIntegers.


Input Format

Input consists of a single integer n


Constraints
1 <= n <= 100


Output Format

Print the factorial of n.



Solution :



title-img


                            Solution in C :

python 3  :

from math import factorial
print(factorial(int(input())))










Java  :

import java.io.*;
import java.util.*;
import java.math.BigInteger;

public class Solution {

    public static void main(String[] args) {
        int N,i;
            Scanner in=new Scanner(System.in);
            N=in.nextInt();
        BigInteger res=BigInteger.ONE;
       for(i=2;i<=N;i++){
       res = res.multiply(BigInteger.valueOf(i));
 
}
        System.out.println(res);
    }
}









C ++  :

#include<iostream>
#include<stdio.h>
using namespace std;

int main(void)
{
    int i=0,j=0,fact[20000],k=0,l=0,n=0,temp=0;

    fact[0]=1;
    l=1;
 
   cin>>n;
    for(i=2;i<=n;i++)
     {
       for(j=0;j<l;j++)
        {k=temp+i*fact[j];
         fact[j]=k%10;
         temp=k/10;
        }
        while(temp>0)
         {
            fact[l++]=temp%10;
            temp=temp/10;
         }
     }
     for(i=l-1;i>-1;i--)
     cout<<fact[i];

   
     return 0;
     }














C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() { int t; int a[200];   int n,i,j,temp,m,x;
   scanf("%d",&n);
   a[0]=1; 
   m=1;    
   temp = 0; 
   for(i=1;i<=n;i++)
   {
        for(j=0;j<m;j++)
        {
           x = a[j]*i+temp; 
           a[j]=x%10; 
           temp = x/10; 
        }
         while(temp>0) 
         { 
           a[m]=temp%10;
           temp = temp/10;
           m++; 
         }
  }
          for(i=m-1;i>=0;i--) 
          printf("%d",a[i]);
          printf("\n");

return 0;
}
                        








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