codeforces 710C Magic Odd Square(构造或者n阶幻方)

Posted 2020-08-04 PrayG

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
分析：给你与1个奇数n, 让你构造一个n * n 的矩阵，要求保证这个矩阵的每行每列和主对角线上数字的和为奇数。
构造：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int ans[50][50];
 4 
 5 int main()
 6 {
 7     int n;
 8     ios::sync_with_stdio(false);
 9     cin.tie(0);
10     cin >> n;
11     memset(ans,0,sizeof(ans));
12     int num1 = 1,num2 =2,cnt1 = (n+1)/2,cnt2 = (n+1)/2;
13     for(int i = 1; i <= n; i++)
14     {
15         for(int j = 1; j <= n; j++)
16         {
17             if(abs(cnt1 - i) + abs(cnt2 - j) <= n/2) // 为什么这么写，有点不清楚
18                 ans[i][j] = num1,num1 += 2;
19             else
20                 ans[i][j] = num2,num2 += 2;
21         }
22     }
23     for(int i = 1; i <= n; i++)
24     {
25         for(int j = 1; j< n; j++)
26         {
27             printf("%d ",ans[i][j]);
28         }
29         printf("%d\n",ans[i][n]);
30     }
31     return 0;
32 }

n阶幻方：

 1 # include <stdio.h>  
 2 # include <string.h>  
 3 int g[50][50];  
 4 int main(){  
 5     int i, j, k, n, x, y, r, c;  
 6     memset(g, 0, sizeof(g));  
 7     scanf("%d", &n);  
 8     r=1; c=(1+n)/2;  
 9     g[1][c]=1;  
10       
11     for(i=2; i<=n*n; i++){  
12         x=r;y=c;  
13         x=x-1;  
14         if(x<1){  
15             x=n;  
16         }  
17         y=y+1;  
18         if(y>n){  
19             y=1;  
20         }  
21         if(g[x][y]){  
22             g[r+1][c]=i;  
23             r=r+1;  
24         }  
25         else{  
26             g[x][y]=i;  
27             r=x;c=y;  
28         }  
29     }  
30     for(i=1; i<=n; i++){  
31         for(j=1; j<=n; j++){  
32             if(j!=n)  
33             printf("%d ", g[i][j]);  
34             else{  
35                 printf("%d\n", g[i][j]);  
36             }  
37         }  
38           
39     }  
40     return 0;  
41 }