[CF710C]Magic Odd Square（构造）

题目链接：http://codeforces.com/contest/710/problem/C

题意：构造一个矩阵，使得行、列、对角线和都是奇数。

可以转换成构造幻方，则每一行、每一列的和为n*(n*n+1)/2。参考http://www.cnblogs.com/codingmylife/archive/2010/12/24/1915728.html 罗伯法

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 59;
int n;
int G[maxn][maxn];

int main() {
    // FRead();
    while(~Rint(n)) {
        Cls(G);
        int i = 0, j = (n - 1) / 2;
        For(k, 1, n*n+1) {
            G[i][j] = k;
            if(k % n == 0) {
                i++;
                continue;
            }
            i--; j++;
            if(i == -1) i += n;
            else if(j > n - 1) j -= n;
        }
        Rep(i, n) {
            Rep(j, n) {
                printf("%d ", G[i][j]);
            }
            printf("
");
        }
    }
    RT 0;
}