打印一个 N*N 的方阵，Ｎ为每边           Ｎ＝１５  打印出下面图形
 字符的个数（３＜Ｎ＜２０）, 要求最               TTTTTTTTTTTTTTT
 外一层为"T", 第二层为"J", 从第三层               TJJJJJJJJJJJJJT
 起每层依次打印数字 1,2,3,...                     TJ11111111111JT
 （右图以Ｎ为１５为例）                           TJ12222222221JT
                                                  TJ12333333321JT
                                                  TJ12344444321JT
                                                  TJ12345554321JT
                                                  TJ12345654321JT
                                                  TJ12345554321JT
                                                  TJ12344444321JT
                                                  TJ12333333321JT
                                                  TJ12222222221JT
                                                  TJ11111111111JT
                                                  TJJJJJJJJJJJJJT
                                                  TTTTTTTTTTTTTTT
#include<stdio.h>
#include<stdlib.h>
//这里主要是运用了规则矩阵的操作
void main()
{
    int i ,j,flag_i,flag_j, N;
    char a[200];
    printf("输入一个数N（3<N<20）");
    scanf("%d",&N);
    if(( N>=20 )||(N<=3))    //主要对输入的数做一定的要求
    {
        printf("输入的数据有问题");
        exit(0);
    }
    for (i = 0;i < (N+1)/2;i++)//利用一个一维数组进行存储这些矩阵
    {
        for(j = 0;j <=i;j++)
        {
            if(j==0)
                a[i*(i-1)+j]='T';
            else
            {
                if(j==1)
                    a[i*(i-1)+j]='J';
                else
                    a[i*(i-1)+j]='0'+j-1;
            }            
        }
    }
   
    for(i=0;i<N;i++)  //按要求对所存储的矩阵进行展开
    {
        flag_i = i;
        for(j=0;j<N;j++)
        {
            flag_j = j;
            if(i>=(N+1)/2)
                i=N-i-1;
            if(j>=(N+1)/2)
                j=N-j-1;
            if(j<=i)
                printf("%c",a[i*(i-1)+j]);
            else
                printf("%c",a[j*(j-1)+i]);
            j=flag_j;
        }
        i=flag_i;
        printf("\n");
    }
}

#include<stdio.h>
char cal(int n, int r, int c)
{
    int d;
    if(n - r - 1 < r) r = n - r - 1;
    if(n - c - 1 < c) c = n - c - 1;
    d = (r < c) ? r : c;
    if(d == 0) return 'T';
    if(d == 1) return 'J';
    return d - 1 + '0';
}
int main()
{
    int n, i, j;
    scanf("%d", &n);
    for(i = 0; i < n; i++, putchar('\n'))
    for(j = 0; j < n; putchar(cal(n, i, j++)));
    return 0;
}

方法挺好的