魔方阵：把1到n*n排成n行n列方阵，使方阵中的每一行、每一列以及对角线上的数之和都相同，即为n阶魔方阵。
根据魔方阵的规律，我将它分为三种情况。
1.奇数阶魔方阵
规律：第一个数放在第一行的中间，下一个数放在上一个数的上一行下一列，若该位置已经有了数字即放在上个数的下面一行的相同列
用C语言编程如下：
示例：n=5;
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>void Magic1()
{
#define ROW 5
#define COL ROW
assert(ROW % 2 != 0); //判断n是否为奇数int arr[ROW][COL] = { 0 }; //定义二维数组int currow = 0;
int curcol = COL / 2;
arr[currow][curcol] = 1;
for (int i = 2; i <= ROW * COL; i++)
{if (arr[(currow - 1 + ROW) % ROW][(curcol + 1) % COL] == 0) //按照规律赋值{currow = (currow - 1 + ROW) % ROW;curcol = (curcol + 1) % COL;}else {currow = (currow + 1) % ROW;}arr[currow][curcol] = i;
}for (int i = 0; i < ROW; i++) //打印魔方阵
{for (int j = 0; j < COL; j++){printf("%-3d", arr[i][j]);}printf("\n");
}}
int main()
{Magic1();return 0;
}结果：
2.偶数阶魔方阵 （n=4K)
规律：按数字从小到大，即1，2，3……n顺序对魔方阵从左到右，从上到下进行填充；
将魔方阵分成若干个4×4子方阵(如：8阶魔方阵可分成四个4×4子方阵)，将子方阵对角线上的元素取出；将取出的元素按从大到小的顺序依次填充到n×n方阵的空缺处。
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
//偶数魔方阵 4K
void Magic2()
{
#define ROW 8
#define COL ROWint tmp = 1;int arr[ROW][COL] = { 0 }; //定义二维矩阵for (int i = 0; i < ROW; i++) {for (int j = 0; j < COL; j++){arr[i][j] = tmp++;}}int row1 = 1;int col1 = 1;int row2 = 1;int col2 = 1;for (int i = 0; i < (ROW / 4) ; i++){for (int j = 0; j < (COL / 4); j++){row1 = 4 * i;col1 = 4 * j;row2 = 4 * i;col2 = 4 * j + 3;for (int k = 0; k < 4; k++){arr[row1][col1] = (ROW * COL + 1) - arr[row1][col1];arr[row2][col2] = (ROW * COL + 1) - arr[row2][col2];row1++;col1++;row2++;col2--;}}}for (int i = 0; i < ROW; i++){for (int j = 0; j < COL; j++){printf("%-3d", arr[i][j]);}printf("\n");}}
int main()
{Magic2();return 0;
}结果：
3.偶数阶魔方阵 （n=4K+2)
规律：
3.1.填充规则
将魔方分成A、B、C、D四个k阶奇方阵， 利用奇数魔方阵填充方法依次将A、D、B、C填充 。
3.2.交换规则 上下标记的数字进行交换
1.右半边大于k+2的列（从1开始）
2.左半边，上下两个块最中心的点进行交换
3.左半边小于中心列的列（除了上下半边最中心的行的第一列的那个值不用交换)(从1开始）
#include<stdio.h>
#include<assert.h>
#include<stdlib.h>void Magic3()
{
#define ROW 10
#define COL ROWassert(ROW % 2 == 0 && ROW % 4 != 0);int arr[ROW][COL] = { 0 };//左上角int currow = 0;int curcol = ROW/4;arr[currow][curcol] = 1;int tmp = 0;for (int i = 2; i <= ROW * COL/ 4; i++){if (arr[(currow - 1 + ROW / 2) % (ROW / 2)][(curcol + 1) % (COL / 2)] == 0) //判断上一行下一列是否被赋值{currow = (currow - 1 + ROW / 2) % (ROW / 2);curcol = (curcol + 1) % (COL / 2);}else{currow = (currow + 1) % (ROW / 2);}arr[currow][curcol] = i;}//右下角currow = ROW / 2;for (int i = 0; i < ROW / 2; i++, currow++){curcol = COL / 2;for (int j = 0; j < COL / 2; j++, curcol++){ arr[currow][curcol] = arr[i][j] + 9;}}//右上角currow = 0;for (int i = ROW/2; i < ROW ; i++, currow++){curcol = COL / 2;for (int j = COL/2; j < COL; j++, curcol++){ arr[currow][curcol] = arr[i][j] + 9;}}//左下角currow = ROW / 2;for (int i = 0; i < ROW/2; i++, currow++){curcol = 0;for (int j = COL/2; j < COL; j++, curcol++){arr[currow][curcol] = arr[i][j] + 9;}}//替换规则1：右半边 大于k+2的列 进行上下交换for (int i = 0; i < ROW / 2; i++){for (int j = ROW / 2 + ROW / 4 + 2; j < COL; j++){tmp = arr[i][j];arr[i][j] = arr[i + ROW / 2][j];arr[i + ROW / 2][j] = tmp;}}//替换规则2：交换左半边，两个中心节点currow = ROW / 4;curcol = COL / 4;tmp = arr[currow][curcol];arr[currow][curcol] = arr[currow + ROW / 2][curcol];arr[currow + ROW / 2][curcol] = tmp;//替换规则3：左半边，除（K+1,1）这个点外，小于k+1的列 上下交换for (int j = 0; j < ROW / 4; j++) //表示交换的列{for (int i = 0; i < ROW / 2; i++) //表示交换的行{if (i == ROW / 4 && j == 0){continue;}else{tmp = arr[i][j];arr[i][j] = arr[i + ROW / 2][j];arr[i + ROW / 2][j] = tmp;}}}//打印for (int i = 0; i < ROW; i++){for (int j = 0; j < COL; j++){printf("%-3d", arr[i][j]);}printf("\n");}
}int main()
{Magic3();return 0;
}结果：
