Răspuns:
1)
nt n,i,j;
cout<<"n=";
cin>>n;
int a[n][n];
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
cout<<"a["<<i<<"]["<<j<<"]=";
cin>>a[i][j];
}
2)
cout<<a[i][j]<<" ";
cout<<endl;
3)
int S=0;
if(i>=j)
S=S+a[i][j];
cout<<S;
4)
int P=1;
if(i+j<=n+1)
P=P*a[i][j];
cout<<P;
5)
if(i<j && a[i][j]>0)
cout<<" "<<a[i][j];
Explicație:
Multumesc+5*+cel mai inteligent raspuns?
