题目链接：http://codeforces.com/contest/710/problem/C
题意：构造一个矩阵，使得行、列、对角线和都是奇数。
可以转换成构造幻方，则每一行、每一列的和为n*(n*n+1)/2。参考http://www.cnblogs.com/codingmylife/archive/2010/12/24/1915728.html 罗伯法
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <cassert> 24 #include <cstdio> 25 #include <bitset> 26 #include <vector> 27 #include <deque> 28 #include <queue> 29 #include <stack> 30 #include <ctime> 31 #include <set> 32 #include <map> 33 #include <cmath> 34 using namespace std; 35 #define fr first 36 #define sc second 37 #define cl clear 38 #define BUG puts("here!!!") 39 #define W(a) while(a--) 40 #define pb(a) push_back(a) 41 #define Rint(a) scanf("%d", &a) 42 #define Rs(a) scanf("%s", a) 43 #define Cin(a) cin >> a 44 #define FRead() freopen("in", "r", stdin) 45 #define FWrite() freopen("out", "w", stdout) 46 #define Rep(i, len) for(int i = 0; i < (len); i++) 47 #define For(i, a, len) for(int i = (a); i < (len); i++) 48 #define Cls(a) memset((a), 0, sizeof(a)) 49 #define Clr(a, x) memset((a), (x), sizeof(a)) 50 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 51 #define lrt rt << 1 52 #define rrt rt << 1 | 1 53 #define pi 3.14159265359 54 #define RT return 55 #define lowbit(x) x & (-x) 56 #define onenum(x) __builtin_popcount(x) 57 typedef long long LL; 58 typedef long double LD; 59 typedef unsigned long long ULL; 60 typedef pair<int, int> pii; 61 typedef pair<string, int> psi; 62 typedef pair<LL, LL> pll; 63 typedef map<string, int> msi; 64 typedef vector<int> vi; 65 typedef vector<LL> vl; 66 typedef vector<vl> vvl; 67 typedef vector<bool> vb; 68 69 const int maxn = 59; 70 int n; 71 int G[maxn][maxn]; 72 73 int main() { 74 // FRead(); 75 while(~Rint(n)) { 76 Cls(G); 77 int i = 0, j = (n - 1) / 2; 78 For(k, 1, n*n+1) { 79 G[i][j] = k; 80 if(k % n == 0) { 81 i++; 82 continue; 83 } 84 i--; j++; 85 if(i == -1) i += n; 86 else if(j > n - 1) j -= n; 87 } 88 Rep(i, n) { 89 Rep(j, n) { 90 printf("%d ", G[i][j]); 91 } 92 printf(" "); 93 } 94 } 95 RT 0; 96 }