FourNArray
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Algorithm Gossip: 4N 魔方阵
说明
与 奇数魔术方阵 相同,在于求各行、各列与各对角线的和相等,而这次方阵的维度是4的倍数。
解法
先来看看4X4方阵的解法: 简单的说,就是一个从左上由1依序开始填,但遇对角线不填,另一个由左上由16开始填,但只填在对角线,再将两个合起来就是解答了;如果N大于2,则以 4X4为单位画对角线: 至于对角线的位置该如何判断,有两个公式,有兴趣的可以画图印证看看,如下所示: 左上至右下:j % 4 == i % 4 右上至左下:(j % 4 + i % 4) == 1
实作
C /#include
/#include /#define N 8 int main(void) { int i, j; int square[N+1][N+1] = {0}; for(j = 1; j <= N; j++) { for(i = 1; i <= N; i++){ if(j % 4 == i % 4 || (j % 4 + i % 4) == 1) square[i][j] = (N+1-i) / N -j + 1; else square[i][j] = (i - 1) / N + j; } } for(i = 1; i <= N; i++) { for(j = 1; j <= N; j++) printf("%2d ", square[i][j]); printf("\n"); } return 0; } Java public class Matrix { public static int[][] magicFourN(int n) { int[][] square = new int[n+1][n+1]; for(int j = 1; j <= n; j++) { for(int i = 1; i <= n; i++){ if(j % 4 == i % 4 || (j % 4 + i % 4) == 1) square[i][j] = (n+1-i) / n -j + 1; else square[i][j] = (i - 1) / n + j; } } int[][] matrix = new int[n][n]; for(int k = 0; k < matrix.length; k++) { for(int l = 0; l < matrix[0].length; l++) { matrix[k][l] = square[k+1][l+1]; } } return matrix; } public static void main(String[] args) { int[][] magic = Matrix.magicFourN(8); for(int k = 0; k < magic.length; k++) { for(int l = 0; l < magic[0].length; l++) { System.out.print(magic[k][l] + " "); } System.out.println(); } } }