# Calculating Factorial (Recursively & Iteratively)

Discussion in 'C' started by pradeep, Oct 30, 2006.

1. ### pr1nc3k1dNew Member

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Did you try the code ? It's working fine ( no overflow ) but it's also limited. I didn't say that it's not. I wanted to write a code which calculates the factorial of 1000 but it's also working for greater values. The vector can hold a result with more than 10.000 digits , but it's limited.

2. ### asadullah.ansariTechCake

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First i told you to store in vector too much overhead and second thing storing has some limit!!! But you know today In Statistical Software factorial of very big number is required. So on that case, By any way you have to go for sterling Approximation formula.

3. ### asadullah.ansariTechCake

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I got the solution for any Big number!!! This code is very efficient and optimized implemented by limked list..
Code:
```class Link
{
//Data Member Parts
private:
int dta;
Link *nxt;
Link *prv;
Link *hd;
Link *rear;

// Function Member Parts

public:
Link(const int& tmp)  :dta(tmp),hd(NULL),rear(NULL),prv(NULL),nxt(NULL)
{
;
}
Link* NxtLnk()   //getting next node
{
return nxt;
}

Link* PrvLnk()  //getting previous node
{
return prv;
}
void ExtraInsert(Link* p);          // After Inserting
int getDta(void)
{
return dta;
}
void SetDta(int temp)
{
dta = temp;
}
int GetNoOfElem();
Link* ToHead();
Link* ToRear();
void DeleteMe(void);
void ClearAll(void);
};

int Link::GetNoOfElem()
{
int count = 0;
Link* p =ToHead();

while(p->NxtLnk()!=NULL){
count++;
p = p->NxtLnk();
}

count++;
return count;
}

void Link::ExtraInsert(Link* p)
{
//    Link* p;
if(prv == NULL) { hd = this;}
p->nxt = nxt;
p->prv = this;
nxt = p;
if(p->nxt == NULL){rear = p;}
};

Link* Link::ToHead()
{
if(hd == this){ //this is the first node
return this;
}

Link *p = this;
while(p->prv != NULL){
p = p->prv;
}

return p;
}

Link* Link::ToRear()
{
if(rear == this){
return this;
}

Link *p = this;
while(p->nxt != NULL){
p = p->nxt;
}

return p;
}

//***** Start ************//
void Link::DeleteMe(void)
{
if(prv == NULL) // First Element
{
nxt->prv = NULL;
hd = nxt;  // Next will be first one
delete this;
return;
}
if(nxt == NULL)
{
prv->nxt = NULL; // last node
rear = prv; //Previous node will be first
return;
}
prv->nxt = nxt;
delete this;
}
//****  End Of Function*******//

void Link::ClearAll(void)
{
Link* p1 = ToHead();
Link* p2;
while(p1->NxtLnk() != NULL){
p2 = p1;
p1 = p1->NxtLnk();
delete p2;
}
delete p1;
};

int main()
{
int n;// remainder
int cy  // Carried over
int rst;
int N;
Link* p = new Link(1);

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

while(p->NxtLnk() != NULL)
{
rst = p->getDta()*n+cy
if(rst>=10){
rn = rst%10;
carry = rst/10;
p->SetDta(rn);
}
else
{
p->SetDta(rst);
}
p = p->NxtLnk();
carry = rst/10;
}
rst = p->getDta()*n+cy

//Other allocated memory for carried over
while(rst >= 10){
Link * newLink = new Link(0);
p->SetDta(rst%10);//rnder
rst = rst/10;
p->ExtraInsert(newLink);
p = p->NxtLnk();
}
p->SetDta(rst);
}

p = p->ToRear();
while(p->PrvLnk()!=NULL)
{
cout<<p->GetDta();
p=p->PrvLnk();
}
cout<<p->GetDta()<<endl;
int num = p->GetNoOfElem();
if(num >=5)
{
p = p->ToRear();
cout<<endl<<"Or"<<endl<<endl;

cout<<p->GetDta()<<".";
p = p->PrvLnk();

for(int i=1;i<5;i++)
{
cout<<p->GetDta();
p = p->PrvLnk();
}
cout<<"E"<num-1<endl;
}
p->ClearAll();
return 0;
}```

Last edited by a moderator: Jan 15, 2008
4. ### shabbirAdministratorStaff Member

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asadullah.ansari, Learn to use Code block when you have code snippets in posts

5. ### pr1nc3k1dNew Member

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Is this code working cuz I tried it and it's not !?

6. ### asadullah.ansariTechCake

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Yes!!! It's working. which plateform you r trying...

7. ### vignesh_svNew Member

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very good job buddy could u help me by saying which is the header file to b included for using
the command sleep( ); in c

8. ### debleena_doll2002New Member

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Why tou are asking this question in this thread. Just create new thread for it in Queries

9. ### lead.smart34New Member

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good info

10. ### crazytolearn57New Member

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good one

11. ### aisha.ansari84New Member

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it takes a lot of time for large numbers