C语言实现魔方阵算法(幻方阵 奇魔方 单偶魔方实现)

#include <stdio.h>
// Author: http://furzoom.com/
// N为魔方阶数
#define N 11
 
int main()
{
    int a[N][N];
    int i;
    int col,row;
 
    col = (N-1)/2;
    row = 0;
 
    a[row][col] = 1;
 
    for(i = 2; i <= N*N; i++)
    {
        if((i-1)%N == 0 )
        {
            row++;
        }
        else
        {
            // if row = 0, then row = N-1, or row = row - 1
            row--;
            row = (row+N)%N;
 
            // if col = N, then col = 0, or col = col + 1
            col ++;
            col %= N;
        }
        a[row][col] = i;
    }
    for(row = 0;row<N;row++)
    {
        for(col = 0;col < N; col ++)
        {
            printf("%6d",a[row][col]);
        }
        printf("\n");
    }
    return 0;
}

#include <stdio.h>
// Author: http://furzoom.com/
// N为魔方阶数
#define N 12
 
int main()
{
    int a[N][N];//存储魔方
    int temparray[N*N/2];//存储取出的元素
    int i;//循环变量
    int col, row;// col 列，row 行
 
    //初始化
        i = 1;
        for(row = 0;row < N; row++)
        {
            for(col = 0;col < N; col ++)
            {
                a[row][col] = i;
                i++;
            }
        }
    //取出子方阵中对角线上的元素，且恰好按从小到大的顺序排放
    i = 0;
    for(row = 0;row < N; row++)
    {
        for(col = 0;col < N; col ++)
        {
             if((col % 4 == row % 4) || ( 3 == ( col % 4 + row % 4)))
            {
                temparray[i] = a[row][col];
                i++;
            }
        }
    }
    //将取出的元素按照从大到小的顺序填充到n×n方阵中
    i = N*N/2 -1;
    for(row = 0;row < N; row++)
    {
        for(col = 0;col < N; col ++)
        {
            if((col % 4 == row % 4) || ( 3 == ( col % 4 + row % 4)))
            {
                a[row][col] = temparray[i];
                i--;
            }
        }
    }
    //输出方阵
    for(row = 0;row < N; row++)
    {
        for(col = 0;col < N; col ++)
        {
            printf("%5d",a[row][col]);
        }
        printf("\n");
    }
    return 0;
}

#include <stdio.h>
// Author: http://furzoom.com/
// N为魔方阶数
#define N 10
 
int main()
{
    int a[N][N] = { {0} };//存储魔方
    int i,k,temp;
    int col,row;// col 列，row 行
 
    //初始化
    k = N / 2;
    col = (k-1)/2;
    row = 0;
    a[row][col] = 1;
    //生成奇魔方A
    for(i = 2; i <= k*k; i++)
    {
        if((i-1)%k == 0 )//前一个数是3的倍数
        {
            row++;
        }
        else
        {
            // if row = 0, then row = N-1, or row = row - 1
            row--;
            row = (row+k)%k;
 
            // if col = N, then col = 0, or col = col + 1
            col ++;
            col %= k;
        }
        a[row][col] = i;
    }
 
    //根据A生成B、C、D魔方
    for(row = 0;row < k; row++)
    {
        for(col = 0;col < k; col ++)
        {
            a[row+k][col+k] = a[row][col] + k*k;
            a[row][col+k] = a[row][col] + 2*k*k;
            a[row+k][col] = a[row][col] + 3*k*k;
        }
    }
 
    // Swap A and C
    for(row = 0;row < k;row++)
    {
        if(row == k / 2)//中间行，交换从中间列向右的m列，N = 2*(2m+1)
        {
            for(col = k / 2; col < k - 1; col++)
            {
                temp = a[row][col];
                a[row][col] = a[row + k][col];
                a[row + k][col] = temp;
            }
        }
        else//其他行，交换从左向右m列,N = 2*(2m+1)
        {
            for(col = 0;col < k / 2;col++)
            {
                temp = a[row][col];
                a[row][col] = a[row + k][col];
                a[row + k][col] = temp;
            }
        }
    }
 
    // Swap B and D
    for(row = 0; row < k;row++)//交换中间列向左m-1列，N = 2*(2m+1)
    {
        for(i = 0;i < (k - 1)/2 - 1;i++)
        {
            temp = a[row][k+ k/2 - i];
            a[row][k+ k /2 -i] = a[row + k][k+k/2 -i];
            a[row + k][k+k/2 -i] = temp;
        }
    }
 
    //输出魔方阵
    for(row = 0;row < N; row++)
    {
        for(col = 0;col < N; col ++)
        {
            printf("%5d",a[row][col]);
        }
        printf("\n");
    }
 
    return 0;
}