CF 710C Magic Odd Square

来源：互联网发布：hyper v linux 编辑：程序博客网时间：2024/05/21 22:29

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

可用n阶奇数幻方求解

奇数幻方：即一个由1~n*n的数字组成的n*n的方阵的每行、列及对角线的数字的和为奇数

奇数阶幻方构造法

(1) 将1放在第一行中间一列;

(2) 从2开始直到n×n止各数依次按下列规则存放：按 45°方向行走，向右上，即每一个数存放的行比前一个数的行数减1，列数加1

(3) 如果行列范围超出矩阵范围，则回绕。例如1在第1行，则2应放在最下一行，列数同样加1;

(4) 如果按上面规则确定的位置上已有数，或上一个数是第1行第n列时，则把下一个数放在上一个数的下面。

n=3时

$/begin{bmatrix}8 & 1 & 6 //3 & 5 & 7 //4 & 9 & 2 ///end{bmatrix}$

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int main(){    int n, i, j, a[100][100], x, y, h, l;    while(cin>>n){        memset(a, 0, sizeof(a));        a[1][n/2+1] = 1;        x = 1;        y = n/2+1;        for(i=2; i<=n*n; i++){            if(x==1)                h = n;            else                h = x-1;            if(y==n)                l = 1;            else                l = y+1;            if(x==1&&y==n){                a[x+1][y] = i;                x = x+1;            }            else if(a[h][l]==0){                a[h][l] = i;                x = h;                y = l;            }            else{                a[x+1][y] = i;                x = x+1;            }        }        for(i=1; i<=n; i++){            for(j=1; j<=n; j++){                if(j==1)                    printf("%d",a[i][j]);                else                    printf(" %d",a[i][j]);            }            cout<<endl;        }    }    return 0;}

0 0