hdu1998 Odd-order Magic (array fill number)

Odd-order Rubik's CubeTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3071 Accepted Submission (s): 1614


Problem description The elements of an n-order phalanx are,..., n^2, each of its rows, each column and 2 diagonal elements and equal, so
Square is called Rubik's Cube. n is an odd number when we have 1 constructs, called "top right", for example, given below n=3,5,7
Rubik's Cube.
3
8 1 6
3 5 7
4 9 2
5
17 24 1) 8 15
23 5 7) 14 16
4 6 13) 20 22
10 12 19) 21 3
11 18 25) 2 9
7
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
The 1th line in the middle of the number is always 1, the last 1 lines in the middle of the number is n^2, his right is 2, from the three Rubik's Cube, you can see "right
Above "What it means."

Input contains multiple sets of data, first entering T, which indicates that there is a T group of data. 1 rows per set of data give N (3<=n<=19) is odd.

Output for each set of data, the N-order Rubik's Cube, 4 squares per number, right-aligned

Sample Input

235

Sample Output

   8   1   6   3   5   7   4   9   2  ,   1   8, 5 7   4 6 (   3  ) of 18   2   9

Author[email protected]
Sourceecjtu Spring Contest
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How to say this question, when I see the first eye of this problem is definitely to find the law, and the entire array is followed.

Because you have to program it. Certainly not too complicated a rule. So it took 5 minutes to find the pattern.

Take n=5 for example, a[0][n/2]=1,a[n-1][n/2]=n*n,a[n-1][n/2+1]=2. And then you'll find 3 in the top right of 2.

5 in the top right of 4 6 is just below 5 ... Wait a minute.. We will find that if M is in a[x][y] position, then m+1 is in the a[x-1][i+1] position (and of course other conditions are satisfied),

As for other conditions that are not satisfied, that is a[x-1][i+1] there are already several, then the m+1 should be under M. And if M is in a[x][n-1], then m+1 is in a[x-1][0]. If M

In A[0][y] then m+1 in a[n-1][y+1]. If M is also below M in a[0][n-1],m+1.

Please attach the code:

#include <stdio.h> #include <string.h>int main () {    int map[20][20],ncase,n;    scanf ("%d", &ncase);    while (ncase--)    {        memset (map,0,sizeof (map));        scanf ("%d", &n);        Map[n-1][n/2]=n*n;        Map[0][n/2]=1;        int t=2;        int i=n-1,j=n/2+1;        while (T<n*n)        {            if (!map[i][j]&&i>=0&&i<n&&j<n&&j>=0)            map[i][j]=t,i--, j++,t++;            else            {                if (i<0&&j<n)                i=n-1;                else if (i>=0&&j==n)                j=0;                else                i+=2,j--;            }        }        for (int i=0;i<n;i++)        {            for (int j=0;j<n;j++)            printf ("%4d", Map[i][j]);            printf ("\ n");}    }    return 0;}

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hdu1998 Odd-order Magic (array fill number)