[C++知识库]魔方阵的实现（三种情况）

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 ROW
	int 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 ROW
	assert(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：?