hello, dear all

```
#include <iostream>
#include <stdio.h>
using namespace std;
int count = 1;
void print( int *arr, int SIZE){
if (arr != 0) {
for (int i = 0; i < SIZE; i++) { // i = position
cout << arr[i];
}
cout <<"\n";
}
}
void circular_left(int *arr, int start, int SIZE)
{
int tmp = arr[start]; //i is a position
for(int i = start; i<= SIZE-1; i++){
arr[i]= arr[i+1];
}
arr[SIZE-1] = tmp;
}
void starterequiv(int *arr, int start, int SIZE)
{ int i, j;
print(arr, SIZE);
if (start < SIZE)
{
for (i = SIZE-1; i > start; i--)
{// outer loop
for (j = i+1 ; j < SIZE; j++)
{ // inner loop
circular_left(arr,i, SIZE);
starterequiv(arr, i, SIZE);
count++;
} //end inner loop
circular_left(arr,i,SIZE);
} // end outer loop
}
}
void initiate(int *arr, int SIZE)
{
for (int i = 0; i <SIZE; i++) { // i is a position
arr[i] = i+1;
}
} // init
int main()
{
int SIZE;
cout << "Enter the number of elements: "; //SIZE = n
cin >> SIZE;
if (SIZE > 0 && SIZE <= 100) {
int *arr = new int[SIZE];
initiate(arr, SIZE);
starterequiv(arr,0,SIZE);
delete [] arr;
cout << "No. Permutation :" << count;
cout << endl;
}
}
```

and then the output as follows:

```
Enter the number of elements: 4
1234
1243
1342
1324
1423
1432
No. Permutation :6
Press any key to continue . . .
```

Actually i need that as my output which at position 1, the element '1' is fixed as a head.

then i just wondering about this routine

```
void starterequiv(int *arr, int start, int SIZE)
{ int i, j;
print(arr, SIZE);
if (start < SIZE)
{
for (i = SIZE-1; i > start; i--)
{// outer loop
for (j = i+1 ; j < SIZE; j++)
{ // inner loop
circular_left(arr,i, SIZE);
starterequiv(arr, i, SIZE);
count++;
} //end inner loop
circular_left(arr,i,SIZE);
} // end outer loop
}
}
```

with the loop 'j'

`for (j = i+1 ; j < SIZE; j++)`

i'm try to throw it but the output is weird. if i left it, i couldn't find where i use 'j' in my algorithm. please help me. i just cycle left all those element exclude '1' to generate next (n-1)! array. it something similar to permutation but unfortunately it is not.