Flipping the Matrix

                            Solution in C :

In C++ :





#include <bits/stdc++.h>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double
#define PI 3.1415926535897932384626

#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second

using namespace std;

int q, n, s[299][299];

int main(){
	scanf("%d", &q);
	while(q--){
		scanf("%d", &n);
		FI(i, 1, 2 * n) FI(j, 1, 2 * n) scanf("%d", &s[i][j]);
		
		int ans = 0;
		FI(i, 1, n) FI(j, 1, n){
			int i2 = n + n + 1 - i;
			int j2 = n + n + 1 - j;
			int mx = max(max(max(s[i][j], s[i][j2]), s[i2][j]), s[i2][j2]);
			
			ans += mx;
		}
		printf("%d\n", ans);
	}
	return 0;
}





In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args)throws IOException {
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(in.readLine());
        for(int t1=0;t1<t;t1++){
            int n=Integer.parseInt(in.readLine());
            String[][] s=new String[2*n][2*n];
            for(int i=0;i<2*n;i++)
                s[i]=in.readLine().split(" ");
            long sum=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    sum+=Math.max(Math.max(Integer.parseInt(s[i][j]),
                                  Integer.parseInt(s[2*n-1-i][j])),
                                  Math.max(Integer.parseInt(s[i][2*n-1-j]),
                                  Integer.parseInt(s[2*n-1-i][2*n-1-j])));
                }
            }
            System.out.println(sum);
        }
    }
}









In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    int q;
    scanf("%d",&q);
    while(q--){
        int n;
        scanf("%d",&n);
        int matrix[2*n][2*n];
        int sol[2*n][2*n];
        for(int i=0;i<2*n;i++){
            for(int j=0;j<2*n;j++){
                scanf("%d",&matrix[i][j]);
                sol[i][j]=0;
            }
        }
        int sum=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                int a = matrix[i][j];
                if(matrix[2*n-1-i][2*n-1-j]>a && sol[2*n-1-i][2*n-1-j]==0){
                    a=matrix[2*n-1-i][2*n-1-j];
                    sol[2*n-1-i][2*n-1-j]++;
                }
                if(matrix[i][2*n-1-j]>a && sol[i][2*n-1-j]==0){
                    a=matrix[i][2*n-1-j];
                    sol[i][2*n-1-j]++;
                }
                if(matrix[2*n-1-i][j]>a && sol[2*n-1-i][j]==0){
                    a=matrix[2*n-1-i][j];
                    sol[2*n-1-i][j]++;
                }
                sum+=a;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}









In Python3 :





q = int(input())
for _ in range(q):
    n = int(input())
    a = []
    for y in range(2*n):
        a.append([int(x) for x in input().split()])
    suma = 0
    for i in range(n):
        for j in range(n):
            suma += max(max(a[i][j],a[2*n-i-1][j]),max(a[i][2*n-j-1],a[2*n-i-1][2*n-j-1]))
    print(suma)