Computer club Question

last year in computer club this question was asked...i am pretty sure gaurav-ji knows the ans...
given a number n.print all the numbers from 1 till n^2 in a spiral.
for example given 3.
print the following
1 2 3
8 9 4
7 6 5

for 4
1 2 3 4
121314 5
111615 6
109 8 7

(i am not able to align for n=4 correctly,but i am sure everyone can get it??? : )

please give logic/code....but mainly tell me the class under which it comes if you know

I started from the last element ie. n^2,then putting all other values from center to first position.
Check this out,if there is some problem in any case then let me know.




#include<iostream>
#include<string>
using namespace std;
#define F0(i,n) for(i=0;i<n;i++)
#define F1(i,n) for(i=2;i<=2*n-1;i++)
int main()
{
int n,ii,jj,put,i,j,k,k1;
cout<<"enter n:";
cin>>n;
put=n*n;
int arr[n][n];
if(n%2==1)
{
ii=1;i=n/2;
jj=-1;j=n/2;
}
else
{
ii=-1;i=n/2;
jj=1;j=n/2-1;
}
arr[i][j]=put;
put--;
F1(k,n)
{
if((k!=2 && k%2==0) || k==2*n-1)
k1=k/2;
else
k1=k/2 + 1;
if(k%2==0)
{
if(k==2)
j=j+jj;
if(k!=2)
i=i+ii;
while(k1--)
{ arr[i][j]=put;
put--;
i=i+ii;
}i=i-ii;
jj*=(-1);
}
else
{ j=j+jj;
while(k1--)
{ arr[i][j]=put;
j=j+jj;
put--;
}
j=j-jj;
ii*=(-1);
}
}
F0(i,n)
{ F0(j,n)
{cout<<arr[i][j]<<" ";}
cout<<"\n";
}
cin>>i;

}

yeah the solution looks fine to me.thanks

i0 = j0 = 0; i1 = m - 1; j1 = n - 1;
for (; {
if (i0 > i1) break;
for (i = i0++, j = j0; j <= j1; j++) a[i][j] = next_input();
if (j0 > j1) break;
for (i = i0, j = j1--; i <= i1; i++) a[i][j] = next_input();
if (i0 > i1) break;
for (i = i1--, j = j1; j >= j0; j--) a[i][j] = next_input();
if (j0 > j1) break;
for (i = i1, j = j0++; i >= i0; i--) a[i][j] = next_input();
}

hey we need to enter the size of the array only and then we have to form the array from that....

it's not next_input() thing.


if u wanna reply ,pls reply with full running(really important) code

int i,j,k,s=1,a[10][10],left,right,up,down,n;
scanf("%d",&n);
left=1;
right=n;
up=2;
down=n-1;
while(s<=n*n)
{for(j=left;j<=right;j++)
a[up-1][j]=s++;
for(i=up;i<=down;i++)
a[i][right]=s++;
for(j=right;j>=left;j--)
a[down+1][j]=s++;
for(i=down;i>=up;i--)
a[i][left=s++;
right=right-1;
left=left+1;
down=down-1;
up=up+1;
}
for(i=1;i<=n;i++)
{printf("\n");
for(j=1;j<=n;j++)
printf("%d",a[i][j]);
}