# 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 :

```                            ```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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