In this post, We will solve HackerRank Extra Long Factorials Problem Solution.
The factorial of the integer n, written n!, is defined as:
n! = n x (n-1) x (n-2) x x3x2 x 1
Calculate and print the factorial of a given integer.
For example, if n = 30, we calculate 30 x 29 x 28 x … x 2 x 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 n > 20 can’t be stored even in a 64-bit 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.
Sample Input
25
Sample Output
15511210043330985984000000
Explanation
25! 25 x 24 x 23 x x 3 x 2 x 1
Extra Long Factorials C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MaxC 200
char O[MaxC];
char S[MaxC];
char T[MaxC];
void Asd(int d1, int d2, int c1, char *o, int *c2) {
int Sum;
Sum = d1 + d2 + c1;
*o = (Sum % 10) + 0x30;
*c2 = (Sum / 10);
}
void Msd(int d1, int d2, int c1, char *o, int *c2) {
int Sum;
Sum = (d1 * d2) + c1;
*o = (Sum % 10) + 0x30;
*c2 = (Sum / 10);
}
void DoFactorial(int i) {
int o, c;
if (i == 100) {
o = 2;
while (o < MaxC-2) {
O[o-2] = O[o];
o++;
}
O[MaxC-3] = '0';
O[MaxC-2] = '0';
}
else {
o = MaxC-2;
c = 0;
while (O[o] != ' ') {
Msd(O[o]-0x30, i%10, c, S+o, &c);
o--;
}
if (c > 0) S[o] = c + 0x30;
if (i>9) {
o = MaxC-2;
c = 0;
while(O[o] != ' ') {
Msd(O[o]-0x30, i/10, c, T+o-1, &c);
o--;
}
if (c > 0) T[o-1] = c + 0x30;
}
o = MaxC-2;
c = 0;
while ((S[o] != ' ') || (T[o] != ' ')) {
if (S[o] == ' ')
Asd(0, T[o]-0x30, c, O+o, &c);
else
if (T[o] == ' ')
Asd(S[o]-0x30, 0, c, O+o, &c);
else
Asd(S[o]-0x30, T[o]-0x30, c, O+o, &c);
o--;
}
if (c > 0) O[o] = c + 0x30;
}
}
int main() {
int n, i;
for (i=0; i<MaxC-2; i++) {
O[i] = ' ';
S[i] = ' ';
T[i] = ' ';
}
O[MaxC-2] = '1';
S[MaxC-2] = '0';
T[MaxC-2] = '0';
O[MaxC-1] = 0;
S[MaxC-1] = 0;
T[MaxC-1] = 0;
scanf("%d", &n);
for (i=1; i<=n; i++) {
DoFactorial(i);
}
i=0;
while (O[i] == ' ') {
i++;
}
printf("%s", O+i);
#if 0
i=0;
while (S[i] == ' ') {
i++;
}
printf("\n%s", S+i);
i=0;
while (T[i] == ' ') {
i++;
}
printf("\n%s", T+i);
#endif
return 0;
}Extra Long Factorials C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int number[300] = {};
int n = 0;
cin >> n;
number[0] = 1;
int len = 1;
for (int i = 2; i <= n; ++i) {
for (int pos = 0, r = 0; pos < len || r != 0; ++pos) {
if (pos < len)
r += number[pos] * i;
else
++len;
number[pos] = r % 10;
r /= 10;
}
}
for (int i = len - 1; i >= 0; --i)
cout << number[i];
return 0;
}Extra Long Factorials C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Linq;
class Solution {
private static string Add(string first, string second)
{
var split = 0;
if (first.Length < second.Length)
{
var s = new StringBuilder();
s.Append('0', second.Length - first.Length);
s.Append(first);
first = s.ToString();
}
if (second.Length < first.Length)
{
var s = new StringBuilder();
s.Append('0', first.Length - second.Length);
s.Append(second);
second = s.ToString();
}
var ret = new StringBuilder();
for (int i = first.Length - 1; i >= 0; i--)
{
var rr = int.Parse(first[i].ToString()) + int.Parse(second[i].ToString()) + split;
split = int.Parse(rr.ToString("00")[0].ToString());
ret.Append(rr.ToString("00")[1]);
}
ret.Append(split);
return new string(ret.ToString().ToCharArray().Reverse().ToArray());
}
private static string mul(string first, string second)
{
var ret = new string('0', first.Length);
for (int i = second.Length - 1; i >= 0; i--)
{
var b = new StringBuilder();
b.Append(mul(first, int.Parse(second[i].ToString())));
b.Append('0', second.Length - 1 - i);
ret = Add(ret, b.ToString());
}
var r = new StringBuilder();
return ret.TrimStart('0');
}
private static string mul(string first, int second)
{
var split = 0;
var ret = new StringBuilder();
for (int i = first.Length-1; i >= 0; i--)
{
var rr = int.Parse(first[i].ToString()) * second+split;
split = int.Parse(rr.ToString("00")[0].ToString());
ret.Append(rr.ToString("00")[1]);
}
ret.Append(split);
return new string(ret.ToString().ToCharArray().Reverse().ToArray());
}
static void Main(String[] args) {
var inp = int.Parse(Console.ReadLine());
string ret = inp.ToString();
for (int i = inp - 1; i > 0; i--)
{
ret = mul(ret, i.ToString());
//Console.WriteLine(ret);
}
Console.WriteLine(ret);
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
}
}Extra Long Factorials Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
import java.math.BigInteger;
class Result {
/*
* Complete the 'extraLongFactorials' function below.
*
* The function accepts INTEGER n as parameter.
*/
public static void extraLongFactorials(int n) {
BigInteger sum=new BigInteger("1");
for(int i=1;i<n+1;i++){
sum=sum.multiply(BigInteger.valueOf(i));
}
System.out.println(sum);
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(bufferedReader.readLine().trim());
Result.extraLongFactorials(n);
bufferedReader.close();
}
}Extra Long Factorials JavaScript Solution
function processData(input) {
//Enter your code here
input
var res = 1
for (var i = 1; i <= input; i++) {
res = mult(res.toString(), i.toString())
}
process.stdout.write(res)
function mult(num1,num2){
var a1 = num1.split("").reverse();
var a2 = num2.split("").reverse();
var aResult = new Array;
for ( iterNum1 = 0; iterNum1 < a1.length; iterNum1++ ) {
for ( iterNum2 = 0; iterNum2 < a2.length; iterNum2++ ) {
idxIter = iterNum1 + iterNum2; // Get the current array position.
aResult[idxIter] = a1[iterNum1] * a2[iterNum2] + ( idxIter >= aResult.length ? 0 : aResult[idxIter] );
if ( aResult[idxIter] > 9 ) { // Carrying
aResult[idxIter + 1] = Math.floor( aResult[idxIter] / 10 ) + ( idxIter + 1 >= aResult.length ? 0 : aResult[idxIter + 1] );
aResult[idxIter] -= Math.floor( aResult[idxIter] / 10 ) * 10;
}
}
}
return aResult.reverse().join("");
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});Extra Long Factorials Python Solution
import math
case = int(input())
print(math.factorial(case))other solutions
