Method of Handling Factorials of Any Size

However, one method to get around this is to dynamically create a linked list such that each element in the list represents a single digit of the result. That way, theorically speaking, the result's length is only limited by the memory!

Steps to Use

  1. Set up a dual_direction linked list,and allocate the first element which is 1 (I'm assuming straight C++. However, if you're using MFC, you can also use an MFC collection class such as the CPtrList class. I personally prefer to create my own class for this purpose.
  2. From the first element to the nth element (last element), simply create an element in the list of each digit until you don't have any more digits in the return result.

Code


//
//
// Nnn.cpp : Defines the entry point for the console application.
// author  : Hai Yi
// date    : Sept,11,2000
//
//

#include "stdafx.h"

#include "iostream.h"
#include "stdlib.h"


//here is a dual link list
class Node{
	
private:
 int data;
 Node *next;
 Node *prev;
 Node *head;
 Node *rear;
	

public:
 Node(const int& item)
 :data(item),prev(NULL),next(NULL),head(NULL),rear(NULL){};

 //get next node
 Node* GetNextNode(){return next;};
 Node* GetPrevNode(){return prev;};

 //insert after
 void InsertAfterMe(Node* p);

 //Delete the appointed
 void DeleteMe(void);

 int GetData(void){return data;};
 void SetData(int item){data = item;};

 //reset
 Node* GoBacktoHead();

 //go to the rear
 Node* GoForwardtoRear();
 //clear the whole
 void ClearAll(void);

 //get the counts of the link
 int GetElementNum();
};


int Node::GetElementNum()
{
 int count = 0;
 Node* p =GoBacktoHead();
 
 while(p->GetNextNode()!=NULL){ 
  count++;
  p = p->GetNextNode();
 }

 count++;
 return count;
}

void Node::InsertAfterMe(Node* p)
{
 //	Node* p;
 if(prev == NULL) { head = this;}
 p->next = next;
 p->prev = this;
 next = p;
 if(p->next == NULL){rear = p;}
};


void Node::DeleteMe(void)
{
 if(prev == NULL) { // if this node is the first one
  next->prev = NULL;
  head = next;  // then the next one becomes the first one
  delete this;  //delete this node
  return;
 }

 if(next == NULL){  //if this node is the last one
  prev->next = NULL;
  rear = prev; // then the previous one becomes the last one
  return;
 }

 prev->next = next;
 delete this;
};

Node* Node::GoBacktoHead()
{
 if(head == this){ //this is the first node
  return this;
 }

 Node *p = this;
 while(p->prev != NULL){
  p = p->prev;
 }

 return p;
}

Node* Node::GoForwardtoRear()
{
 if(rear == this){
  return this;
 }

 Node *p = this;
 while(p->next != NULL){
  p = p->next;
 }

 return p;
}

void Node::ClearAll(void)
{
 Node* p = GoBacktoHead();
 Node* p2;
 while(p->GetNextNode() != NULL){
  p2 = p;
  p = p->GetNextNode();
  delete p2;
 }

 delete p;
};

int main(int argc, char* argv[])
{
 int remain;
 int carry;
 int result;
 int N;
 Node* p = new Node(1);

 cout<<"Pls input the number:";
 cin>>N;
 for(int n=1;n<=N;n++)
 {
  remain = carry = 0;
  p = p->GoBacktoHead();

  //while not the end of the list,process the element one by one
  while(p->GetNextNode() != NULL){
   result = p->GetData()*n+carry;
   if(result>=10){
    remain = result%10;
    carry = result/10;
    p->SetData(remain);
   }
   else{p->SetData(result);}

  p = p->GetNextNode();
  carry = result/10;
  }

  result = p->GetData()*n+carry;
		
  //if carry occurs,process the carry and 
  //store into the newly allocated space.

  while(result >= 10){
   Node * newNode = new Node(0);
   p->SetData(result%10);//remainder
   result = result/10;
   p->InsertAfterMe(newNode);
   p = p->GetNextNode();
  }

  p->SetData(result);

 }//end of if

 p = p->GoForwardtoRear();

 while(p->GetPrevNode()!=NULL){
  cout<<p->GetData();
  p=p->GetPrevNode();
 }

 cout<<p->GetData()<<endl;
 int num = p->GetElementNum();
 if(num >=5){
  p = p->GoForwardtoRear();

  cout<<endl<<"Or"<<endl<<endl;

  cout<<p->GetData()<<".";
  p = p->GetPrevNode();

  for(int i=1;i<5;i++){
   cout<<p->GetData();
   p = p->GetPrevNode();
  }

  cout<<"E"<num-1<endl;
 }

 //clear the memory
 p->ClearAll();

 return 0;
}


Comments

  • cheap dr dre kd ndy tfcm wr dxxbw aitiaw yybzf

    Posted by carpinteyrowar on 03/14/2013 08:28am

    qk qddth ybudvr crdby yvi nudu th cgnte atdfvh khcqx six zbkj vh ainsu wkihsv pscfa jqt ohae h Our updates Recent articles: http://www.kokonnya.com/archives/2008/12/fedora10web.html#comments http://christianradionewsroom.org/outreach/blog1.php/holidays/ http://www.yamakan.or.jp/furukawa/2012/10/post-12.html#comments

    Reply
  • JKXCNE PA JT SOP usDG Ex

    Posted by eHRzFovMca on 01/27/2013 05:20am

    generic viagra online purchase of viagra in us - viagra 50 dosage

    Reply
Leave a Comment
  • Your email address will not be published. All fields are required.

Top White Papers and Webcasts

  • Live Event Date: August 14, 2014 @ 2:00 p.m. ET / 11:00 a.m. PT Data protection has long been considered "overhead" by many organizations in the past, many chalking it up to an insurance policy or an extended warranty you may never use. The realities of today make data protection a must-have, as we live in a data driven society. The digital assets we create, share, and collaborate with others on must be managed and protected for many purposes. Check out this upcoming eSeminar and join eVault Chief Technology …

  • Hybrid cloud platforms need to think in terms of sweet spots when it comes to application platform interface (API) integration. Cloud Velocity has taken a unique approach to tight integration with the API sweet spot; enough to support the agility of physical and virtual apps, including multi-tier environments and databases, while reducing capital and operating costs. Read this case study to learn how a global-level Fortune 1000 company was able to deploy an entire 6+ TB Oracle eCommerce stack in Amazon Web …

Most Popular Programming Stories

More for Developers

Latest Developer Headlines

RSS Feeds