CodeForces 710C Magic Odd Square (n阶奇幻方)

题意：给它定一个n，让你输出一个n*n的矩阵，使得整个矩阵，每行，每列，对角线和都是奇数。

析：这个题可以用n阶奇幻方来解决，当然也可以不用，如果不懂，请看：http://www.cnblogs.com/dwtfukgv/articles/5797527.html

剩下的就很简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 100000000000000000;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline LL Max(LL a, LL b){  return a < b ? b : a; }
inline LL Min(LL a, LL b){  return a > b ? b : a; }
inline int Max(int a, int b){  return a < b ? b : a; }
inline int Min(int a, int b){  return a > b ? b : a; }int a[55][55];
int main(){
    while(scanf("%d", &n) == 1){
        int j = n/2, num = 1, i = 0;
        while(num != n*n + 1){
            int ii = (i % n + n) % n;
            int jj = (j % n + n) % n;
            a[ii][jj] = num;
            if(num % n == 0)  ++i;
            else --i, ++j;
            ++num;
        }
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                if(j == n-1)   printf("%d\n", a[i][j]);
                else printf("%d ", a[i][j]);
    }
    return 0;
}