How to print numbers in a spiral order?

Thank you , i am trying to solve a project euler problem it wants me to print the sum of

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

this is formed by starting with the number 1 and moving to the right in a clockwise direction for a 5 by 5 matrix but i am in trouble writing a code for the spiral matrix !!

Solution

It is Highly recommended to do project Euler problems on your own and ask for help if you are really stuck

here is how i will write a code in c to print a spiral as suggested in the question

#include<stdio.h>
main()
{
int i,j,nq=9;//nq is a odd number which represents the order of the matrix
int lim=(int)nq/2,cnt=2;
int a[nq][nq];
 for(i=0;i<nq;i++){
    for(j=0;j<nq;j++)
    a[i][j]=0;
    }
a[lim][lim]=1;
a[lim][lim+1]=2;
int i1=lim,j1=lim+1;i=lim,j=lim;
while(1){
       if(cnt>(nq*nq))
         break;
        cnt++;
if(i==i1)
{  j=j1;
   if(i<=lim)
   {
       i=i1;
    if(a[i1+1][j1]==0)
        a[++i1][j]=cnt;
    else
       a[i1][++j1]=cnt;
   }
   else
   {   i=i1;
    if(a[i1-1][j1]==0)
       a[--i1][j1]=cnt;
    else
        a[i1][--j1]=cnt;
   }
}
else
{   i=i1;
    if(j<lim)
   {
        j=j1;
       if(a[i1][j+1]==0)
        a[i1][++j1]=cnt;
       else
        a[--i1][j1]=cnt;
   }
   else
   {    j=j1;
       if(a[i1][j1-1]==0)
        a[i1][--j1]=cnt;
       else
        a[++i1][j1]=cnt;
   }
  }
}
for(i=0;i<nq;i++){
    for(j=0;j<nq;j++)
    printf(" %d    ",a[i][j]);
    printf("\n");
}

}

I Googled your question http://projecteuler.net/problem=28 this can also be solved by taking advantage of its mathematical nature note that

Top right corner is n^2 and other corners can be shown to be n^2-2n+2 ,n^2-n+1, and n^2-3n+3. you just need to sum those corners which comes to be

= 4*n^2 - 6*n + 6

hence the final answer can be calculated by iterating over every second number from 1001 to 3

long int sum(int n){
    long int sum=1;
    while(n>1){
      sum=sum+4*n*n-6*n+6;
      n=n-2;
    }
    return sum;
    }