Trying to do some tricky char output

I'm doing a problem from a c++ book and am lost and clueless.... the jist of the problem is that the user enters a char from the keyboard then it prints the letters out in a triangle starting from the inside&last row first, for instance if the char was d the out put would look like
      A
    ABA
  ABCBA
ABCDCBA

I'v figured that that a for loop with 2 nested loops (one for spacing, and one for lines) are needed, at least I think........
if anyone could point me in the right direction on how to implement them, and get the output

thanks for the help.......

-orlando-
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Orlando15767Asked:
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n_fortynineCommented:
This sounds like homework, so here's a few hints:
Yes, two nested loops seem sufficient. The outer should be used to control the "row" of the triangle (e.g. if the letter is 'D' how many rows are there? how about 'E'? what's the relationship here?)

The inner loop is used to displayed each row. Notice that you can start from 'A' going to 'D' and coming back to A, so something like:
    Say A + n is the letter (in case letter is 'D' n is 3)
(A) (A+1) (A+2) ... (A + (n - 1)) (A+n) (A + (n - 1)) ... (A + 2) (A + 1) (A)
Everything revolves around n, pretty much. Remember that the center letter varies for each row (so does n), and this is what your outer loop should be keeping track off.
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freewellCommented:
int i,j,nDepth = 4;
char ch = 'A';

for(i=0;i<nDepth;i++)
{
      for(j=1;j<nDepth-i;j++)
            printf(" ");
      for(j=ch;j<=ch+i;j++)
            printf("%c",j);
      for(j=ch+i-1;j>=ch;j--)
            printf("%c",j);

      printf("\r\n");
}
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Orlando15767Author Commented:
how can you figure out what the position of the alphabet the letters are in (d=5,E=6) to figure out the number of lines and rows?
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n_fortynineCommented:
Well, the simplest way to find that out:
ch = toupper(ch); //make everything to upper case (or to lower and then subtract 'a'.
int pos = ch - 'A';
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Orlando15767Author Commented:
Below is the code that I came up with( I put the '-' in to see where a space would go.......
there is a few problems with it though, the output is all on one line, and I can't get the letters to decrease.....  any suggestions?

this is what the out put looks like...... '- - - DDD - - DD - D'

 int i,j;
      char letter;

      cout << "Enter a char to start the triangle: "; cin >> letter;

      letter = toupper(letter);

      int pos = (letter - 'A')+1;

      cout << letter << "'s position in the alphebet is: " << pos << endl;
      cout << "\n" << endl;

      for ( ; pos >= 1; pos--)
      {
      for (i=1; i<pos; i++)
            cout << "-";
      for (j=1; j<pos; j++)
            cout << letter;
      }

thanks for the help
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n_fortynineCommented:
I'm not going to write any code here, since freewell has already given you some code that would work for you (sigh). I'll just explain what he did in case you didn't understand it. (If you use his code please give him credit for it)

for(i=0;i<nDepth;i++)
{
     for(j=1;j<nDepth-i;j++) //Display the preceding spaces here.
          printf(" ");                //Notice that you would go to nDepth - 1, meaning this limit would change every time.
                                         //and reaches 0 on the last row.
     for(j=ch;j<=ch+i;j++)    //Printing the first segment of the row: A, AB, ABC etc. The limit would also change.
          printf("%c",j);
     for(j=ch+i-1;j>=ch;j--)  //Printing the remaining segment of the row.
          printf("%c",j);

     printf("\r\n");
}

You could replace printf() with the corresponding C++ syntax cout.
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Orlando15767Author Commented:
I tried his code and it doesn't work how I thought it would, and wound't use his anyway... I want to work it though myself....
such as his code doesn't work if you change the ch to 'z'
in his code what does "%c" mean and such......


thanks for all the help with this....
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n_fortynineCommented:
well, the thing is you need to calculate nDepth the way you did, i.e. if the letter is 'D' nDepth is 4 etc.
for all his C syntax:
printf(" "); is equivalent to cout << ' ';
printf("%c",j); is equivalent to cout << j; ("%c" saying this variable is of type char)
printf("\r\n"); is equivalent to cout << endl;

If you want to work it through by yourself, well, you can still break the task into 3 (or 2 if you try to compress the last 2) smaller portions (per row):
1. Display the preceding spaces.
2. Display the first half segment.
3. Display the rest.

Have fun.
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Orlando15767Author Commented:
this is what I'v got so from peicing this together. the thing is it is outputing numbers instead of letters
thanks for all the help


int i,j;
char ch;

      cout << "Enter a char to start the triangle: "; cin >> ch;

      ch = toupper(ch);

      int pos = (ch - 'A')+1;

for(i=0;i<pos;i++)
{
    for(j=1;j<pos-i;j++)
         cout << "-";
    for(j=ch;j<=ch+i;j++)
      {
            //char j;
            cout << j;
      }
    for(j=ch+i-1;j>=ch;j--)
      {
            //char j;
            cout << j;
      }
    cout << endl;
}
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