1.(比較無聊的循環(huán))Change/add only one character and print '*' exactly 20 times.
(there are atleast 3 solutions to this problem :-)
int main()
{
 int i, n = 20;
 for (i = 0; i < n; i--)
  printf("*");
 return 0;
}

解1：
int main()
{
 int i, n = 20;
 for (i = 0; -i < n; i--)
  printf("*");
 return 0;
}

解2：
int main()
{
 int i, n = 20;
 for (i = 0; i < n;n--)
  printf("*");
 return 0;
}

解3：
int main()
{
 int i, n = 20;
 for (i = 0; i + n; i--)
  printf("*");
 return 0;
}

2. display below like:
        1
       212
      32123
     4321234
    543212345

解：(雙重循環(huán))
for (int i = 1; i <= n; i++)
{
    for (int j = 2 * n - 1; j >= 1; j--)
    {
        if (Math.Abs(j - n) >= i)
              Console.Write(' ');
        else
              Console.Write(Math.Abs(j - n) + 1);
    }
    Console.WriteLine();
}