UVa problem solution 11470 - Square Sums

problem link:

Discuss: if you know the 2D. then the problem may pretty for you. how i did it first calculate all corner index value.then just increase the value of i and j. and you may trouble how to choose the corner point as it change then use them and analyse  ara[i][j], ara[j][n-1-i], ara[n-1-i][n-1-j], ara[n-1-j][i]  first traverse till n/2 and second loop traverse till n-1-i; examine it . it traverse all the square of the grid. then just print it..


try yourself, before see the code




#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,cas=0;
    while(cin >>n&&n)
    {
        cas++;
        int i,j,cnt=0,flag=0,ara[11][11];
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                cin >>ara[i][j];
            }
        }
        cout <<"Case "<<cas<<':';
        for(i=0;i<n/2;i++)
        {

            int sum=0;
            for(j=i;j<n-1-i;j++)
            {
                sum+=ara[i][j]+ara[j][n-1-i]+ara[n-1-i][n-1-j]+ara[n-1-j][i];
            }
            cout <<' '<<sum;
        }
        if(n%2)
            cout <<' '<<ara[n/2][n/2];
            cout <<endl;
    }
    return 0;
}