3. 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                                                  TTTTTTTTTTTTTTTint main(void){  char c,a[8]={'T','J','1','2','3','4','5','6'};      int i,j;  for(i=0;i<15;i++){      for(j=0;j<15;j++){          if(i<=8){            if(j<=i)              c=a[j];            else if(14-j<=i) c=a[14-j];          }          else{             if(j<=14-i)               c=a[j];             else if(14-j<=14-i) c=a[14-j];           }           printf("%c",c);           if(j==14) printf("\n");    } }   return 0;         }/*4. 在Ｎ行Ｎ列的数阵中, 数Ｋ（１〈＝Ｋ〈＝Ｎ）在每行和每列中出现且仅  出现一次，这样的数阵叫Ｎ阶拉丁方阵。例如下图就是一个五阶拉丁方阵。  编一程序，从键盘输入Ｎ值后，打印出所有不同的Ｎ阶拉丁方阵，并统计个数。        1  2  3  4  5        2  3  4  5  1        3  4  5  1  2        4  5  1  2  3        5  1  2  3  4*/int main(void){  int i,j,k;  printf("请输入一个比26小的数字:\n");  scanf("%d",&k);  if(k>25){      printf("数值太大，打印不下，程序退出。\n");      exit(0);  }  for(i=0;i<k;i++){      for(j=1;j<=k;j++){        (j+i<=k)?printf("%3d",j+i):printf("%3d",i+j-k);        if(j==k) printf("\n");    }  }  return 0;}

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