Thread: For finding the factorial of a large number

Here is my algorithm :-
You know to calculate factorial of a number, you should multiply all numbers from 1 to itself, for example:

http://cboard.cprogramming.com/1000-...iles/minus.gif Collapse | Copy Code
[IMG]file:///images/minus.gif[/IMG] Collapse | Copy Code

1000! = 1*2*3*……*998*999*1000 There is no variable in any programming language that can store the result of this multiplication exactly; except storing in scientific notation. Because of sequential multiplication's huge result, finding a new strategy or mechanism is necessary.In my recommended algorithm, the number is not looked at as a number, instead it is treated as sequential digits where each one has its own numerical value. In fact, an array of digits is available. Each time, the second numberis multiplied by each digit of the first number (index of array) and then added with carry number up from the previous multiplication.
This process is shown, for example 12!:

http://cboard.cprogramming.com/1000-...iles/minus.gif Collapse | Copy Code
[IMG]file:///images/minus.gif[/IMG] Collapse | Copy Code

12! = 11! * 12 11! = 39916800 12! = 479001600 http://cboard.cprogramming.com/1000-...es/article.jpg

• If the result is less than 10, the result will be placed in the cell of array.
• If it is equal to or greater than 10, it will be divided by 10 and then the remainder will be placed in the array's cell and quotient placed in variable that will be added with next multiplication's response.
  Note: Notice that all the numbers are stored from end of array, the same as normal calculation (real calculation).

Code:

#include<stdio.h>
  2 int main()
  3 {
  4         int t,count,n,p,count1,i,carry=0,total,k=0;
  5         int array[500];
  6         scanf("%d",&t);
  7         for(count=1;count<=t;count++)
  8         {
  9                 scanf("%d",&n);
 10                 array[499]=1;
 11                 if(n>1)
 12                 {
 13                         printf("*");
 14                         for(count1=2;count1<=n;count++)
 15                         {
 16                                 carry=0;
 17                                 for(i=0;i<n;i++)
 18                                 {
 19                                         array[499-i]=(array[499-i]*count1)+carry;
 20                                         if(array[499-i]>9)
 21                                         {
 22                                                 carry=array[499-i]/10;
 23                                                 array[499-i]=array[499-i]%10;
 24                                                 if(count1==n)
 25                                                         k++;
 26                                         }
 27                                         else if(array[499-i]<=9)
 28                                         {
 29                                                 carry=0;
 30                                                 if(count1==n)
 31                                                         k++;
 32                                         }
 33                                 }
 34                         }
 35                 }
 36                 if(n==1)
 37                         k=1;
 38                 for(p=500-k;p<500;p++)
 39         //      for(p=475;p<500;p++)
 40                         printf("*%d",array[500-k]);
      printf("*");
 42         }
 43         return 0;
 44 }

Nothing is printing..... ...I m a beginner....please help

By the way, people have done exactly this already, and made nice libraries for it.

For example: The GNU MP Bignum Library

Originally Posted by ShubhamDepp

Nothing is printing..... ...I m a beginner....please help

As suggested above, you ALWAYS want to work with a simple example of the problem, and THEN go on to the larger problem.


Any run-time constraints facing this program? What is the expected range of N, where N! is what the program is calculating?

Your links were not working for me. I'll look it over tomorrow, but I expect you'll have worked out the bugs by then, anyway.

I am interested though, in what you expect for this program.