// Course:            CS 12

// Lecture Section:   1
// Lab Section:       21
// Assignment #:      In-lab exercise 1

// First Name:        Wagner
// Last Name:         Truppel
// ID Number:         12-345-678
// Email address:     wagner@cs.ucr.edu

// =======================================================================

#include <iostream>

void f(const unsigned char c, const unsigned int n);

int main ()
{
    unsigned char c;
    std::cout << "Please enter the character to be used: \n";
    std::cin >> c;
    
    unsigned int n;
    std::cout << "Please enter the number of lines per branch, n: \n";
    std::cin >> n;
    
    unsigned int h;
    std::cout << "Please enter the number of tree branches, h: \n";
    std::cin >> h;
    
    // print h branches
    for (unsigned int i = 0; i < h; i++)
    {
        f(c, n);
    }
    
    return 0;
}

// f(c, n) prints n lines, where the top one has a single
// appearance of the character c, the second one has three
// appearances of c, the third one has five appearances of
// c, and so on, all of them centered.
//
// The key idea is to notice the patterns. If the 1st
// line has 1 character, the 2nd has 3, the 3rd has 5,
// and so on, then the n-th line will have (2*n - 1)
// characters. Also, the last line [the n-th line] has
// no blank spaces, the next-to-last [the (n-1)-th line]
// has 1 blank at the start and 1 at the end, the (n-2)-th
// line has 2 blanks at the start and 2 at the end, and so
// on. Thus, the first line has (n-1) blanks at the start
// and (n-1) blanks at the end. We can ignore the blanks
// at the end; we only need to print the blanks at the
// start and the correct number of characters.
//
// To summarize, the k-th line must have (n-k) blanks and
// then (2*k - 1) copies of the character c. We need to
// print n such lines.
void f(const unsigned char c, const unsigned int n)
{
    for (unsigned int k = 1; k <= n; k++)
    {
        // print (n-k) blanks
        for (unsigned int r = 0; r < (n - k); r++)
            std::cout << " ";
        
        // print (2*k - 1) c's
        for (unsigned int r = 0; r < (2*k - 1); r++)
            std::cout << c;
        
        // print a newline character at the end of each line
        std::cout << std::endl;
    }
}