Find an n × n matrix with different numbers from 1 to n², so the sum in each row, column and both main diagonals are odd.

The only line contains odd integer n (1 ≤ n ≤ 49).

Print n lines with n integers. All the integers should be different and from 1 to n2. The sum in each row, column and both main diagonals should be odd.

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <stack>

#include <vector>

#include <queue>

#include <cstring>

#include <string>

#include <map>

#include <set>

using namespace std;

const int BufferSize = 1 << 16;

char buffer[BufferSize], *Head, *Tail;

inline char Getchar() {

    if(Head == Tail) {

        int l = fread(buffer, 1, BufferSize, stdin);

        Tail = (Head = buffer) + l;

    }

    return *Head++;

}

int read() {

    int x = 0, f = 1; char c = getchar();

    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

    return x * f;

}

#define maxn 55

int n, A[maxn][maxn];

int main() {

	n = read();

	int x = 1, y = (n >> 1) + 1, tot = 0;

	A[x][y] = ++tot;

	while(tot < n * n) {

		if(x == 1 && y != n){ x = n; y++; A[x][y] = ++tot; continue; }

		if(x != 1 && y == n){ x--; y = 1; A[x][y] = ++tot; continue; }

		if(x == 1 && y == n){ x++; A[x][y] = ++tot; continue; }

		if(x != 1 && y != n) {

			if(!A[x-1][y+1]){ x--; y++; A[x][y] = ++tot; continue; }

			x++; A[x][y] = ++tot;

		}

	}

	for(int i = 1; i <= n; i++) {

		for(int j = 1; j < n; j++) printf("%d ", A[i][j]);

		printf("%d\n", A[i][n]);

	}

	return 0;

}