Educational Codeforces Round 16 -- C - Magic Odd Square （找规律）

大体题意：

给你一个奇数n，求出一个n*n矩阵，使得n*n矩阵  每一行，每一列，和主对角线上的数字之和都是奇数！

思路：

这个题出的很晚了，刚开始一直在推n 是偶数怎么办，没有看到n 是奇数= =！

又没看到要求对角线上也是奇数，交了发Wa!没想到最后还能改出来！

其实这个题目分析下样例就可以看出规律了！

他每一行奇数的个数是 1 ， 3， 5 ，7 ，，n ，，7 ， 5 ， 3， 1.

把这些奇数放在中间，偶数从2开始填，奇数从1开始填就可以了！

#include<bits/stdc++.h>using namespace std;const int maxn = 300000 + 10;const int inf = 0x3f3f3f3f;const double eps = 1e-10;const double pi = acos(-1.0);typedef long long ll;int a[50][50];int vis[2500];int hang[50];int main(){    int n;    scanf("%d",&n);    if (n == 1){        printf("1\n");        return 0;    }    int cnt = 1;    for (int i = 1; i <= (n+1)/2; ++i){        hang[cnt++] = i*2-1;    }    for (int i = n/2; i >= 1; --i){        hang[cnt++] = i*2-1;    }    --cnt;    int cur = 0;        int curji = 1,curou = 2;    for (;;){        if (curji >= n && curou >= n-1)break;        for (int i = 1; i <= n; ++i){                int j;            for (j = 1; j<= (n-hang[i])/2; j++ )a[i][j] = curou,curou += 2;            int curr = j-1;            for (; j <= curr + hang[i]; ++j)a[i][j]=curji,curji+=2;            for (;j <= n; ++j)a[i][j] = curou,curou+=2;        }    }    for (int i = 1; i<= n; ++i){        for (int j = 1; j <= n; ++j){            if (j > 1)printf(" ");            printf("%d",a[i][j]);        }        puts("");    }    return 0;}

C. Magic Odd Square

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output

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.

Examples

input

1

output

1

input

3

output

2 1 43 5 76 9 8