c How to output a diamond

Suitable for students to ask some C language topics, this would like to search the Internet, but online methods are divided into the upper and lower parts with two nested for loop output. I think there is no combination of output, so that in line with the requirements of simple and beautiful? The answer is yes.

Title: The output of the row is equal to the diamond, such as 5 rows 5 columns:

Ranks 12345

1 *

2 * * *

3 * * * *

4 * * *

5 *

2. Analysis:

   When the number of rows equals half of the total row number, the ' * ' is incremented and incremented to 2.

   When the number of rows is greater than half the total row number, the ' * ' is decremented and incremented to 2.

   It has a diamond with a row n,

   We use the variable I as the number of rows, starting with 0, I<n.

   The number of ' * ' to be exported by K as the first line of N-Ling.

   Then the median value of the i<=n (in M, m= (n-1)/2), K increments, and when i>m, K decreases.

   Because K is symmetric relative to M line, so:

   When I>m, the K value for each increment of i-m is equal to the K value of line m-i, i.e. I-m line and M (i-m) are relative.

   So we get the result:

   When I<=m, k=2i+1

   When I>m, k=2 (M (i-m)) +1, (m= (n-1)/2). namely K=2n-2i-1

   Since we only need to output the space on the left of ' * ', the number of spaces to the left of each line ' * ' is: (n per line of K)/2.

   Each line loops out the space, then loops out the ' * ' of each line, and then outputs a carriage return, ending a row.

   So we can start programming.

4.  

   #include <stdio.h>

   int main ()

   {

    int i,j,k,n;//defines four variables: I is row, J number of spaces before each line *, K the * numbers for each line

    printf ("Please input odd n:");

    scanf ("%d", &n);  //enter odd n

    for (i=0;i<n;i++)  //loop all rows

    {

     if (n>=2*i+1)  //determine K value

       k=2*i+1;

     Else

       k=2*n-2*i-1;

     for (j= (n-k)/2;j>0;j--)  //Circular output spaces

       printf ("");

     for (; k>0;k--)  //Circular output * Number

       printf ("*");

     printf ("n");  //output carriage return end line

   }

    

6.  

   The title is over, but the above procedure is not perfect

   For example, what if someone enters an n value for an even number? Enter a non-numeric value?

   We add something to go in:

   #include <stdio.h>

    

   int main ()

   {

   int i,j,k,n=0;   This gives an initial value to N, even if the input is Non-numeric, n can also have an exact number.

   do{           //Add a large loop that allows N to be entered multiple times to observe different results.

   printf ("Please input odd N, 0 to exit:")///when entering 0, exit the large loop and end the program.

   scanf ("%d", &n);

   if (!) ( n%2)    //increase the judgment of the input even number.

   {

   printf ("You are input a even, please input AGAIN.N");

   Continue;

   }

   for (i=0;i<n;i++)

   {

   if (n>=2*i+1)

   K=2*i+1;

   Else

   K=2*n-2*i-1;

   J= (n-k)/2;

   for (; j>0;j--)

   printf ("");

   for (; k>0;k--)

   printf ("*");

   printf ("n");

   }

   }while (n);

   Return 0;

   }