C语言魔方阵的三种实现方法
时间:2022-03-05 09:50:38|栏目:C代码|点击: 次
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:
上一篇:C++实现LeetCode(208.实现字典树(前缀树))
栏 目:C代码
本文标题:C语言魔方阵的三种实现方法
本文地址:http://www.codeinn.net/misctech/195250.html