Question

i came across the following program for calculating large factorials(numbers as big as 100).. can anyone explain me the basic idea used in this algorithm??
I need to know just the mathematics implemented in calculating the factorial.

#include <cmath>
#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{

      unsigned int d;

      unsigned char *a;

      unsigned int j, n, q, z, t;

      int i,arr[101],f;

      double p;


    cin>>n;
    p = 0.0;
    for(j = 2; j <= n; j++)
        p += log10(j);
    d = (int)p + 1;
    a = new unsigned char[d];
    for (i = 1; i < d; i++)
        a[i] = 0; //initialize
    a[0] = 1;
    p = 0.0;
    for (j = 2; j <= n; j++)
    {
        q = 0;
        p += log10(j);
        z = (int)p + 1;
        for (i = 0; i <= z/*NUMDIGITS*/; i++)
        {
            t = (a[i] * j) + q;
            q = (t / 10);
            a[i] = (char)(t % 10);
        }

    }
    for( i = d -1; i >= 0; i--)
        cout << (int)a[i];
    cout<<"\n";
    delete []a;

return 0;
}

Answer 1

Note that

n! = 2 * 3 * ... * n

so that

log(n!) = log(2 * 3 * ... * n) = log(2) + log(3) + ... + log(n)

This is important because if k is a positive integer then the ceiling of log(k) is the number of digits in the base-10 representation of k. Thus, these lines of code are counting the number of digits in n!.

p = 0.0;
for(j = 2; j <= n; j++)
    p += log10(j);
d = (int)p + 1;

Then, these lines of code allocate space to hold the digits of n!:

a = new unsigned char[d];
for (i = 1; i < d; i++)
    a[i] = 0; //initialize

Then we just do the grade-school multiplication algorithm

p = 0.0;
for (j = 2; j <= n; j++) {
    q = 0;
    p += log10(j);
    z = (int)p + 1;
    for (i = 0; i <= z/*NUMDIGITS*/; i++) {
        t = (a[i] * j) + q;
        q = (t / 10);
        a[i] = (char)(t % 10);
    }
}

The outer loop is running from j from 2 to n because at each step we will multiply the current result represented by the digits in a by j. The inner loop is the grade-school multiplication algorithm wherein we multiply each digit by j and carry the result into q if necessary.

The p = 0.0 before the nested loop and the p += log10(j) inside the loop just keep track of the number of digits in the answer so far.

Incidentally, I think there is a bug in this part of the program. The loop condition should be i < z not i <= z otherwise we will be writing past the end of a when z == d which will happen for sure when j == n. Thus replace

for (i = 0; i <= z/*NUMDIGITS*/; i++)

by

for (i = 0; i < z/*NUMDIGITS*/; i++)

Then we just print out the digits

for( i = d -1; i >= 0; i--)
    cout << (int)a[i];
cout<<"\n";

and free the allocated memory

delete []a;