very very helpful post.
–Thanks
2 Likes
A lot of…lot of thanks to u 
2 Likes
Hell yeah!!! …I use python and VOILA!!!
2 Likes
what is t here?didn’t understand y while(t–) is used?pls explain…
1 Like
I have a better solution not just for finding factorial.
Whenever the question involves computation with big numbers (yeah very big!)
you can use a user defined data type BIGINT you need not do any code for it I’m posting it here(Even I found it somewhere )
hope you find it useful.
here’s the code:
#include <iostream>
#include
#include
#include
#include <string.h>
#include <math.h>
using namespace std;
const int base = 1000000000;
const int base_digits = 9;
struct bigint {
vector a;
int sign;
bigint() :
sign(1) {
}
bigint(long long v) {
*this = v;
}
bigint(const string &s) {
read(s);
}
void operator=(const bigint &v) {
sign = v.sign;
a = v.a;
}
void operator=(long long v) {
sign = 1;
if (v < 0)
sign = -1, v = -v;
for (; v > 0; v = v / base)
a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int) max(a.size(), v.a.size()) || carry; ++i) {
if (i == (int) res.a.size())
res.a.push_back(0);
res.a[i] += carry + (i < (int) a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry)
res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int) v.a.size() || carry; ++i) {
res.a[i] -= carry + (i < (int) v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry)
res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {
if (i == (int) a.size())
a.push_back(0);
long long cur = a[i] * (long long) v + carry;
carry = (int) (cur / base);
a[i] = (int) (cur % base);
//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((long long) base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0)
r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const {
return divmod(*this, v).first;
}
bigint operator%(const bigint &v) const {
return divmod(*this, v).second;
}
void operator/=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = (int) a.size() - 1, rem = 0; i >= 0; --i) {
long long cur = a[i] + rem * (long long) base;
a[i] = (int) (cur / v);
rem = (int) (cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0)
v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (long long) base) % v;
return m * sign;
}
void operator+=(const bigint &v) {
*this = *this + v;
}
void operator-=(const bigint &v) {
*this = *this - v;
}
void operator*=(const bigint &v) {
*this = *this * v;
}
void operator/=(const bigint &v) {
*this = *this / v;
}
bool operator<(const bigint &v) const {
if (sign != v.sign)
return sign < v.sign;
if (a.size() != v.a.size())
return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i])
return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const {
return v < *this;
}
bool operator<=(const bigint &v) const {
return !(v < *this);
}
bool operator>=(const bigint &v) const {
return !(*this < v);
}
bool operator==(const bigint &v) const {
return !(*this < v) && !(v < *this);
}
bool operator!=(const bigint &v) const {
return *this < v || v < *this;
}
void trim() {
while (!a.empty() && !a.back())
a.pop_back();
if (a.empty())
sign = 1;
}
bool isZero() const {
return a.empty() || (a.size() == 1 && !a[0]);
}
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
long long longValue() const {
long long res = 0;
for (int i = a.size() - 1; i >= 0; i--)
res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) {
return b.isZero() ? a : gcd(b, a % b);
}
friend bigint lcm(const bigint &a, const bigint &b) {
return a / gcd(a, b) * b;
}
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-')
sign = -sign;
++pos;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++)
x = x * 10 + s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream& operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream& operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1)
stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int) v.a.size() - 2; i >= 0; --i)
stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<long long> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int) p.size(); i++)
p[i] = p[i - 1] * 10;
vector<int> res;
long long cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int) a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int) cur);
while (!res.empty() && !res.back())
res.pop_back();
return res;
}
typedef vector<long long> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++)
a2[i] += a1[i];
for (int i = 0; i < k; i++)
b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int) a1b1.size(); i++)
r[i] -= a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
r[i] -= a2b2[i];
for (int i = 0; i < (int) r.size(); i++)
res[i + k] += r[i];
for (int i = 0; i < (int) a1b1.size(); i++)
res[i] += a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a6 = convert_base(this->a, base_digits, 6);
vector<int> b6 = convert_base(v.a, base_digits, 6);
vll a(a6.begin(), a6.end());
vll b(b6.begin(), b6.end());
while (a.size() < b.size())
a.push_back(0);
while (b.size() < a.size())
b.push_back(0);
while (a.size() & (a.size() - 1))
a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int) c.size(); i++) {
long long cur = c[i] + carry;
res.a.push_back((int) (cur % 1000000));
carry = (int) (cur / 1000000);
}
res.a = convert_base(res.a, 6, base_digits);
res.trim();
return res;
}
};
int main()
{
int t;
cin>>t;
bigint fact=1;
for(int i=1;i<=t;i++)
fact*=i;
cout<<fact<<endl;
return 0;
}
3 Likes
thanks for this very nice information
I’ve tried to get the factorials by the same method using a different approach but i cannot get factorials of numbers above 30. Can you please tell me where i did it wrong? Here is my code
#include <iostream>
using namespace std;
int main()
{
int i,j,x,n,a[2000],m=1,fac=1,temp=0;
cout << "Enter no. to calculate factorial"<<endl;
cin >> n;
a[0]=1;
for(i=1;i<=n;i++)
{
for(j=0;j<m;j++)
{
if(j==0)
temp=0;
x=i*a[j]+temp;
a[j]=x%10;
temp=x/10;
}
if(temp>0)
{
a[m]=temp;
m++;
}
}
cout<<"Factorial of " <<n;
cout<< " is ";
for(i=m-1;i>=0;i--)
{
cout<<a[i];
}
}Hello all,
It is a great technique to store factorial in an array.But how to use it in calculation.Suppose we need to calculate combination of n and r(nCr) and say n is 1000 and r is 100. Then how to use these factorial stored in array.
JAVA:
import java.util.Scanner;
import java.math.BigInteger;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int T;
T=sc.nextInt();
while(T>0){
int n, c;
BigInteger inc = new BigInteger(“1”);
BigInteger fact = new BigInteger(“1”);
n = sc.nextInt();
for (c = 1; c <= n; c++) {
fact = fact.multiply(inc);
inc = inc.add(BigInteger.ONE);
}
System.out.println(fact);
T–;
}
}
}
There is one more solution . Use long double and while printing write cout.precision(200)<<answer;
#include
#include
using namespace std;
long double fact(long double num){
if(num==1)
return 1;
else{
return num* fact(num-1);
}
}
int main(int argc, char * argv[]){
int T;
cin>>T;
while(T–){
long double num,ans;
cin>>num;
cout.precision(200);
ans=fact(num);
cout<<ans;
cout<<’\n’;
}
return 0;
}
// This solution gives the same answer. gcc4.8.1
Thanks I learned this now hehe.
Thanks a lot, a very very well explained topic. This tutorial really helped…!!
So I had anyways, written my own after reading this post; difference is that I have used char array(char[]). My program can easily calculate factorial up to 15000. 15000! is calculated in 2.00 sec to 2.07 seconds(according to Ideone).
I believe it can further be optimized. I am looking for a code which lets me easily calculate (10^9)! My search is on. One thought I got is to trim the trailing zeros, after 900 or 1000. This can significantly reduce calculations. While printing the result, the number of appending zeros can be calculated from a different way. 15000! contains something 3700 trailing zeros.
Anyways, I welcome comments on my code. Here is my code:
#include <stdio.h>
char res[100000];
int main() {
int n,i,j,m;
long temp,c;
while(scanf("%d",&n)!=EOF)
{
m=1;
res[0]=‘1’;
for(i=2;i<=n;i++){
c=0;
for(j=0;j<m;j++){
temp=((res[j]-48)*i)+c;
res[j]=(temp%10)+48;
c=temp/10;
}
while(c>0)
{
res[m]=(c%10)+48;
c=c/10;
m++;
}
}
for(i=m-1;i>=0;i–)
printf("%c",res[i]);
printf("\n");
}
return 0;
}
Please do suggest how to improve this code…!!
omg, this is really helpful. Thank you.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace BigMultiplier
{
class Program
{
static void Main(string[] args)
{
int[] s1 = new int[1];
int[] s2 = new int[1];
s1[0]=1;
Program p = new Program();
int[] s3 = p.doit(s1, s2);
Console.WriteLine("Enter the Number for Which Factorial to be Found (below 1000)");
int limit = Convert.ToInt32(Console.ReadLine());
for (int i = 1; i <= limit; i++)
{
s3 = p.return_array(i);
s1 = p.doit(s1, s3);
}
int sum = 0;
for (int j = s1.Length-1; j >=0; j--)
{
Console.Write(s1[j]);
sum = sum + s1[j];
}
Console.Write("sum = "+sum);
Console.WriteLine();
Console.ReadLine();
}
int[] return_array(int num)
{
int[] num_arr = new int[1]; ;
if (num <= 9)
{
num_arr = new int[1];
num_arr[0] = num;
}
else if (num <= 99)
{
num_arr = new int[2];
num_arr[1] = num % 10;
num = num / 10;
num_arr[0] = num;
}
else if (num <= 999)
{
num_arr = new int[3];
num_arr[2] = num % 10;
num = num / 10;
num_arr[1] = num % 10;
num = num / 10;
num_arr[0] = num % 10;
}
else if (num <= 9999)
{
num_arr = new int[4];
num_arr[3] = num % 10;
num = num / 10;
num_arr[2] = num % 10;
num = num / 10;
num_arr[1] = num % 10;
num = num / 10;
num_arr[0] = num % 10;
}
else if (num <= 99999)
{
num_arr = new int[5];
num_arr[4] = num % 10;
num = num / 10;
num_arr[3] = num % 10;
num = num / 10;
num_arr[2] = num % 10;
num = num / 10;
num_arr[1] = num % 10;
num = num / 10;
num_arr[0] = num % 10;
}
return num_arr;
}
int[] doit(int[] num1_int, int[] num2_int)
{
// String num1_string, num2_string;
// Console.WriteLine(“Enter the Number 1”);
// num1_string = Console.ReadLine();
// Console.WriteLine(“Enter the Number 2”);
// num2_string = Console.ReadLine();
// int[] num1_int = new int[num1_string.Length];
// int[] num2_int = new int[num2_string.Length];
// for (int j = 0; j < num1_string.Length; j++)
// {
// num1_int[j] = num1_string[j]-48;
// Console.Write(" " + num1_int[j]);
// }
// for(int j=0;j<num2_string.Length;j++)
// {
// num2_int[j] = num2_string[j]-48;
// Console.Write(" " + num2_int[j]);
// }
int[,] num3_int = new int[num2_int.Length, (num1_int.Length + 1)];
int i,k,temp=0;
//Multiplication on Individual Digits Done and the Values are there in the Individual Cells of the Array
for(i=0;i<num2_int.Length;i++)
{
for (k = 0; k < num1_int.Length; k++)
{
int mul = (num1_int[k] * num2_int[i])+temp;
num3_int[i, k] = mul % 10;
temp = mul / 10;
if (k == (num1_int.Length - 1))
{
num3_int[i, k+1] = temp;
}
// Console.Write(" " + num3_int[i, k]);
}
// Console.Write(" " + num3_int[i, k]);
temp=0;
// Console.WriteLine();
}
// temp = 0;
// Console.ReadLine();
int[] result_int = new int[num1_int.Length + num2_int.Length];
// int[,] num3_int = new int[num2_string.Length, (num1_string.Length + 1)];
double result=0;
for(i=0;i<num2_int.Length;i++)
for (k = 0; k < (num1_int.Length + 1); k++)
{
// Console.Write(" i = " + i + " k=" + k+" “);
// Console.Write(num3_int[i, k]);
result=result+(num3_int[i,k]*Math.Pow(10,k)*Math.Pow(10,i));
// Console.WriteLine(” "+(num3_int[i,k]*Math.Pow(10,k)*Math.Pow(10,i)));
// Console.WriteLine("10^k " + Math.Pow(10,k));
}
int[,] num4_int= new int[num2_int.Length,num1_int.Length + num2_int.Length];
for (i = 0; i < num2_int.Length; i++)
{
for (k = 0; k < (num1_int.Length + 1); k++)
{
//for (int l = 0; l < 0; l++)
{
num4_int[i,k + i] = num3_int[i,k];
}
}
}
for (i = 0; i < num2_int.Length; i++)
{
for (k = 0; k < (num1_int.Length + num2_int.Length); k++)
{
// Console.Write(" " + num4_int[i, k]);
}
// Console.WriteLine();
}
int[] re_int = new int[num1_int.Length + num2_int.Length];
temp=0;
for (i = 0; i < re_int.Length; i++)
{
re_int[i] = 0;
for (k = 0; k < num2_int.Length; k++)
{
re_int[i] = num4_int[k, i] + re_int[i];
}
int t = re_int[i] + temp;
re_int[i]=t%10;
temp=t/10;
if(i==(re_int.Length-1))
{
//Need to check
}
}
// Console.WriteLine("Final Result - Reversed Order “);
// for(i=0;i<re_int.Length;i++)
// Console.Write(” "+re_int[i]);
// re_int[i+k]=;
// Console.WriteLine("Final Result - Correct Order “);
// for(i=re_int.Length-1;i>=0;i–)
// Console.Write(” " + re_int[i]);
// Console.WriteLine(result);
// Console.ReadLine();
return re_int;
}
}
}
ok thats a good idea
Thanks a lot
I was always confused to do this task. And finally your post removed my confusion.
Thank you for your effective article.
The efficient way to calculate factorial is by calculating k-th last element on k-th iteration. If you can output to a file all you need is O(n) space lesser multiplications. infact, we can do that without multiplications.
Checkout for full article: