// [10/11/2014 Sjm]
/*
(数组下标默认从 1 开始)
经分析可知：
令 myNode = ((n + 1) >> 1)，
则： myArr[myNode][myNode] = 1^2
myArr[myNode + 1][myNode + 1] = (1 + 2)^2
myArr[myNode + 2][myNode + 2] = (3 + 2)^2
…………
据此可模拟生成过程。。。
*/
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <cmath>
5 #include <algorithm>
6 #include <functional>
7 #include <string>
8 #include <cstring>
9 #include <vector>
10 #include <stack>
11 #include <queue>
12 #include <map>
13 using namespace std;
14 #define eps 1e-8
15 #define MAX_LEN 102
16
17 int n;
18 int myArr[MAX_LEN][MAX_LEN];
19
20 void Solve() {
21 int val = 1;
22 int myNode = ((n + 1) >> 1);
23 myArr[myNode][myNode] = val * val;
24 for (int i = myNode + 1; i <= n; ++i) {
25 val += 2;
26 myArr[i][i] = val * val;
27 int northwest = myNode - (i - myNode);
28 for (int j = 1; j < val; ++j) {
29 myArr[i][i - j] = myArr[i][i] - j;
30 }
31 for (int j = 1; j < val; ++j) {
32 myArr[i - j][northwest] = myArr[i][northwest] - j;
33 }
34 for (int j = 1; j < val; ++j) {
35 myArr[northwest][northwest + j] = myArr[northwest][northwest] - j;
36 }
37 for (int j = 1; j < (val - 1); ++j) {
38 myArr[northwest + j][i] = myArr[northwest][i] - j;
39 }
40 }
41 int Sum = 0;
42 for (int i = 1; i <= n; ++i) {
43 for (int j = 1; j <= n; ++j) {
44 printf("%d", myArr[i][j]);
45 if (j != n) printf(" ");
46 else printf("
");
47 if ((i == j) || (i + j == n + 1)) {
48 Sum += myArr[i][j];
49 }
50 }
51 }
52 printf("%d
", Sum);
53 }
54
55 int main() {
56 scanf("%d", &n);
57 Solve();
58 return 0;
59 }