codeforces 710C C. Magic Odd Square(构造)

题目链接：

C. Magic Odd Square

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 4
3 5 7
6 9 8

题意:

找出一个n*n的矩阵,这里面的数是一个[1,n*n]的全排列,要求所有行所有列的和为奇数;n为奇数;

思路：

可以发现n为奇数的时候就是每行每列的个数都是奇数,所有每行每列里面的奇数的个数都是奇数,所有就可以想到中间是一个奇数组成的45度的正方形,其他是偶数;正好用完了所有的偶数和奇数;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=3e5+10;
const int maxn=1e3+20;
const double eps=1e-12;

int ans[50][50];

int main()
{

    int n;
    read(n);

        int cnt1=1,cnt2=2,cx=n/2+1,cy=n/2+1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(abs(cx-i)+abs(cy-j)<=n/2)ans[i][j]=cnt1,cnt1+=2;
                else ans[i][j]=cnt2,cnt2+=2;
            }
        }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<n;j++)printf("%d ",ans[i][j]);
        printf("%d\n",ans[i][n]);
    }

    return 0;
}