1.#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= ");

       scanf("%d",&n);

       k=1;

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++) //每一層輸入 i 個(gè)元素，故j<=i.

              {

              a[i+1-j][j] = k; //在內(nèi)循環(huán)，并且與j有關(guān)。

//第i層的第一個(gè)數(shù)的橫坐標(biāo)為i+1-j,然后隨著j++而橫坐標(biāo)漸漸斜上輸入，

                        //數(shù)組第二個(gè)下標(biāo)以 j 作為基值，隨著j++縱坐標(biāo)向右延伸。

                     k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n+1-i;j++) //每一行輸出 n+1-i 個(gè)元素

                            printf("%4d",a[i][j]);

              }

       printf("\n");

}

2 代碼：與上一個(gè)程序相比只僅僅改變了：a[i+1-j][j] —> a[j][i+1-j].

#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= "); 

  scanf("%d",&n);

k=1;

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++)

              {

                     if(i%2==0)   a[i+1-j][j]=k;

                  else         a[j][i+1-j]=k;           

                     k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n+1-i;j++)

                  printf("%4d",a[i][j]);

              } 

     printf("\n"); 

 }

5、代碼：

#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= ");

       scanf("%d",&n);

       k=1;

//輸出上三角

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++)

              {

                     if(i%2==0)  a[i+1-j][j]=k;

                  else        a[j][i+1-j]=k;          

                     k++;

              }

//輸出下三角

     for(i=n+1;i<2*n;i++)

               for(j=1;j<=2*n-i;j++)

          {

                     if(i%2==0)  a[n+1-j][i-n+j]=k;   //n+1在這里是定值了

                  else        a[i-n+j][n+1-j]=k;       

                     k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n;j++)

                            printf("%4d",a[i][j]);

              }

       printf("\n");

}

6、代碼：只是把上面5代碼中兩個(gè)地方交換位置

#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= ");

       scanf("%d",&n);

       k=1;

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++)

              {

                     if(i%2==0)  a[j][i+1-j]=k;

                  else        a[i+1-j][j]=k;         

                     k++;

              }

     for(i=n+1;i<2*n;i++)

               for(j=1;j<=2*n-i;j++)

          {

                     if(i%2==0)   a[i-n+j][n+1-j]=k;

                  else         a[n+1-j][i-n+j]=k;      

                     k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n;j++)

                     printf("%4d",a[i][j]);

              }

       printf("\n");

}

#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= ");

       scanf("%d",&n);

       k=1;

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++)          

{

                     a[j][i+1-j]=k;        

       k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n+1-i;j++) //每一行輸出 n+1-i 個(gè)元素

                            printf("%4d",a[i][j]);

              }

       printf("\n");

}

3、代碼：與上一個(gè)相比加了一個(gè)奇偶數(shù)層的if判斷，這正好就是“拐點(diǎn)”的關(guān)鍵！

#include<stdio.h>

void main()

{

       int i,j,a[100][100],n,k;

       printf("請(qǐng)輸入列數(shù)n= ");

       scanf("%d",&n);

       k=1;

       for(i=1;i<=n;i++)

              for(j=1;j<=i;j++)

              {

                     if(i%2==0)  a[j][i+1-j]=k;

                  else        a[i+1-j][j]=k;          

                     k++;

              }

              for(i=1;i<=n;i++)

              {

                     printf("\n");

                     for(j=1;j<=n+1-i;j++)

                            printf("%4d",a[i][j]);

              }

       printf("\n");

}