Presupunand ca problema este de programare raspunsul ar arata cam asa:
for(k=1;k<=(nrlin+1)/2;k++)
{
for(j=k;j<=nrlin-k+1;j++)
cout<<a[k][j]<<" ";
for(i=k+1;i<=nrlin-k+1;i++)
cout<<a[i][nrlin-k+1]<<" ";
for(i=nrlin-k;i>=k;i--)
cout<<a[nrlin-k+1][i]<<" ";
for(i=nrlin-k; i>k;i--)
cout<<a[i][k]<<" ";
}