Question

我试着编写一个打印带有镜像字母的钻石的程序，如下面的形状：

       a
      aba
     abcba
    abcdcba
   abcdedcba
  abcdefedcba
   abcdedcba
    abcdcba
     abcba
      aba
       a

这是我已经做过的事情：

 #include <stdio.h>

int main()
{
  int n, c, k,f=0, space = 1;
  char ch='a';

  printf("enter the size of diamond\n");
  scanf("%d", &n);

  space = n - 1;

  for (k = 1; k <= n; k++)
  {
    for (c = 1; c <= space; c++)
      printf(" ");

    space--;


    for (c = 1; c <= 2*k-1; c++)

     if (c <= f)
     printf("%c", ch);
     ch++;
     if (c>f)
     ch--;
     printf("%c", ch);
     printf("\n");
  }







  space = 1;

  for (k = 1; k <= n - 1; k++)
  {
    for (c = 1; c <= space; c++)
      printf(" ");

    space++;

    for (c = 1 ; c <= 2*(n-k)-1; c++)
      printf("a");

    printf("\n");
  }

  return 0;
}

但显然这段代码中有一些错误，任何人都可以帮我检测一下吗？

Answer 1

int n, c, k, space;

printf("enter the size of diamond\n");
scanf("%d", &n);

char s[n+2];
s[0] = 'a';s[1] = '\0';

for(c = 0; c < n; ++c){
    space = n - c -1;
    printf("%*s", space, "");
    k = printf("%s", s) - 1;
    while(--k >=0)
        putchar(s[k]);
    puts("");
    s[c+1] = s[c] + 1;
    s[c+2] = '\0';
}
for(c = 1; c < n; ++c){
    space = c;
    printf("%*s", space, "");
    k = printf("%.*s", n - c, s) - 1;
    while(--k >=0)
        putchar(s[k]);
    puts("");
}

#include <stdio.h>

void print(char from, char to){
    putchar(from);
    if(from < to){
        print(from+1, to);
        putchar(from);
    }
}

int main(void){
    int c = 0, n, k = 1;

    printf("enter the size of diamond\n");
    scanf("%d", &n);

    do{
        printf("%*s", n - c -1, "");
        print('a', 'a' + c);
        puts("");
        if(c + 1 == n)
            k = -k;
        c += k;
    }while(c>=0);

    return 0;
}

Answer 2

如果替换此

for (c = 1 ; c <= 2*(n-k)-1; c++)
    printf("a");

用这个

for (c = 0 ; c < 2*(n-k)-1; c++)        // note the different conditions
    printf("%c", 'a' + c);              // print char instead of string

你应该得到正确长度的每一行，但增加的字符如

abcde

但它并没有从中点减少。我把它留给你了！

Answer 3

将任务分开运行会更容易。因此，根据位置和线长计算要打印的字符的功能是：

int getCharAtPos(int pos, int len)
{
    if (pos > len / 2)
    {
        return 'a' + len - pos;
    }
    else
    {
        return 'a' + pos - 1;
    }
}

然后使用

而不是printf("a")

    for (c = 1; c <= 2*k-1; c++)
      printf("%c", getCharAtPos(c, 2*k-1));

.....
    for (c = 1 ; c <= 2*(n-k)-1; c++)
      printf("%c", getCharAtPos(c, 2*(n-k)-1));

Answer 4

问题的第二个解决方案。你可以使用递归。我为你的任务写了一个例子。

//--------------
 // Recursive Out one Simbol
 void DamSimRec(char chOut ,int scnt)
  {
        printf("%c", chOut);


        if (scnt > 1)
        {
            DamSimRec(chOut + 1, scnt - 1);
            printf("%c", chOut);
        }
}// Print Space
//--------------
    void SpaceOut(int pSizeSpace)
    {
        int a_c;
        for (a_c = 0; a_c < pSizeSpace; a_c++)
            printf(" ");
    }
//--------------
 // Recursive Print One String Krylov E.A.
  void DamRec(int space, int sout)
    {

        SpaceOut(space);//space;

        DamSimRec('a', sout);//Simbols

        printf("\n");

        if (space > 0)
            DamRec(space-1, sout+1);

        if (sout > 1)

            SpaceOut(space + 1);
            DamSimRec('a', sout - 1);
        }
        printf("\n"); 

    }
main()
{
        int aSize;
        printf("enter the size of diamond\n");
        scanf_s("%d", &aSize);

        DamRec(aSize , 1);

}

...所以，你可以使用它，但要记住堆栈。

使用镜像字母在C中打印钻石

4 个答案: