Go4Expert - All possible combinations of a string

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All possible combinations of a string

All possible combinations of a string
1. Finds all possible cominations of a given string. The total number of combinations with recurrences is the factorial of the string length.
2. Have given the word 'post' as an example since if I included a cin >> word; then a user may unconscientiously enter a long word which will take toooo long to end. To change to another word change the string str.
3. A 9 letter word like 'aeroplane' can take around 40 seconds on a 1GHz machine. A 10 letter word will take 9 x 40 s - thats 6 mins! So watch out for the string length.
4. This may time according to the length of the string str

Code: CPP

// Combination

#include<iostream.h>
#include<string.h>
#include<conio.h>

unsigned long fact(unsigned long); // finds factorial of a number
void swap(int[],int,int);
void display(int[],char[]);
void Add(int[],int);

void main()
{
    // clrscr(); // Remove the first comment tag if running in TC++ or BC++ DOS mode
    int j,k,n;
    
    unsigned long i;
    char str[]="post"; // small word
    n=strlen(str);
    int*key=new int[n+1];
    key[0]=n; // key[0] contains the total no: of letters
    for (i=1;i<=key[0];i++)
        key[i]=i;
    display(key,str);
    swap(key,key[0],key[0]-1);
    display(key,str);
    for (i=3;i<=fact(key[0]);i+=2)
    {
        for (j=key[0]-1;j>=0;j--)
        {
            if ((i-1)%fact(j)==0)
            {
                Add(key,key[0]-j);
                for (k=1;k<=key[0]-j-1;k++)
                    if (key[k]==key[key[0]-j])
                    {Add(key,key[0]-j);k=0;}
            }
        }
        display(key,str);
        swap(key,key[0],key[0]-1);
        display(key,str);
    }
    cout << "\nTotal no: of combinations = " << key[0] << "! = " << i-1;
    getch();
}

unsigned long fact(unsigned long f)
{ 
    return f==0?1:f*fact(f-1); 
}

void swap(int k[],int a,int b)
{
    int t=k[a];k[a]=k[b];k[b]=t;
}

void display(int k[],char s[])
{
    for(int i=1;i<=k[0];i++)
        cout << s[k[i]-1];
    cout << '\t';
}

void Add(int k[],int i)
{
    k[i]++;
    if (k[i]==k[0]+1)
        k[i]=1;
}