题干：
C语言_魔方阵
描述
魔方阵是一个古老的智力问题，它要求在一个m×m的矩阵中填入1～m2的数字（m为奇数），使得每一行、每一列、每条对角线的累加和都相等，如下为5阶魔方阵示例。
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
输入
输入一个数n，表示矩阵的大小，保证n<100且n为奇数
输出
输出一个矩阵，每个元素之间以一个空格分隔
答案不唯一，输出一个题目要求的矩阵即可
输入样例 1
5输出样例 1
15 8 1 24 1716 14 7 5 2322 20 13 6 43 21 1912 109 2 25 18 11
解题报告：
可以找到规律构造，以（1，n/2+1）这个点为1，然后向左上扩展，如果最上面了就循环到最下面，如果到最左边了就循环到最右边，如果该点被填过值了，那就到他下面那个点，然后就可以构造出这个图形了。
刚开始是用搜索写的：
#include<bits/stdc++.h>using namespace std;int n,flag,sum;bool vis[505];int maze[505][505];bool fitr1(int x) {int res = 0;for(int i = 1; i<=n; i++) {res += maze[x][i];}if(res != sum) return 0;else return 1;}bool fitr2(int x ,int y){int res = 0;for(int i = 1; i<=y; i++) {res += maze[x][i];}if(res >= sum) return 0;else return 1;}bool dui(int x,int y) {int res = 0;for(int i = 1; i<=x; i++) {res += maze[i][i];}if(res >sum ) return 0;res = 0;for(int i = 1; i<=x; i++) {res += maze[i][n-i+1];}if(res >sum ) return 0;return 1;}bool fitc(int x,int y) {int res = 0;for(int i = 1; i<=x; i++) {res += maze[i][y];}if(res > sum) return 0;else return 1;}bool ff() {int resr[105] = {0},resc[105] = {0},res = 0;for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {resr[i] += maze[i][j];resc[j] += maze[i][j];}}for(int i = 1; i<=n; i++) {if(resr[i] != sum || resc[i] != sum) return 0;}//for(int i = 1; i<=n; i++) res += maze[i][i];if(res != sum ) return 0;res = 0;for(int i = 1; i<=n; i++) res += maze[i][n-i+1];if(res != sum ) return 0;return 1;}void dfs(int x,int y) {//printf("x = %d y = %d\n",x,y);if(x == n+1) {if(ff()) flag = 1;return ;}if(flag == 1) return ;for(int i = 1; i<=n*n; i++) {if(vis[i]) continue;maze[x][y] = i;if(flag == 1) return ;if(x == y) {if(!dui(x,y)) continue;}//if(!fitc(x,y)) continue;if(y == n) {//if(!fitr1(x)) continue;vis[i] = 1;dfs(x+1,1);if(flag == 1) return ;vis[i] = 0;}else {//if(!fitr2(x,y)) continue;vis[i]=1;dfs(x,y+1);if(flag == 1) return ;vis[i] = 0;}}}int main(){cin>>n;for(int i = 1; i<=n*n; i++) {sum += i;}sum/=n;dfs(1,1);for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {printf("%d%c",maze[i][j],j==n ? '\n' : ' ');}}return 0 ;}然而发现只能输入3的时候输出，输入5的时候就跑不出来了。最后一项也是，9^15次方，肯定跑不出来呀。
AC代码：
#include<cstdio>#include<queue>#include<string>#include<cstring>#include<cmath>#include<map>#include<iostream>#include<algorithm>#define ll long longconst ll mod = 1e9+7;using namespace std;int maze[105][105];int main(){int n;cin>>n;maze[1][(n+1)/2] = 1;int cur=2,tmp = n;int x = 1,y = (n+1)/2;while(cur <= n*n) {if(x == 1) {if(y == 1) {maze[x+1][y] = cur++;x++;tmp = n;continue;}else {for(int i = 1; i<=n; i++) {if(maze[tmp][y-1] != 0) tmp--;else break;}maze[tmp][y-1] = cur++;x=tmp;y--;tmp = n;}}else if(y == 1) {for(int i = 1; i<=n; i++) {if(maze[x-1][tmp] != 0) tmp--;else break;}maze[x-1][tmp] = cur++;x--;y=tmp;tmp=n;}else if(maze[x-1][y-1] != 0 ) {maze[x+1][y] = cur++;x++;}else {maze[x-1][y-1] = cur++;x--,y--;}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {printf("%d%c",maze[i][j],j == n ? '\n' : ' ');}}return 0 ;}
