#include <iostream>

// Query the user for one positive integer 'm'
// and draw a radical (square-root) symbol whose
// dimensions depend on 'm'.
//
// If m = 5, the radical symbol should appear
// as follows:
//
//                        <------15----->
//                        ***************
//                       * ^
//                      *  |
//                     *   5
//                    *    |
//                   *     v
//  ^ **            *
//  |   **         *
//  5     **      *
//  |       **   *
//  v         ***
//
// The labels and arrows above are shown only to
// illustrate the dimensions. Your program should
// print only the asterisks and spaces.
//
// The horizontal bar contains 3*m asterisks. The
// upper diagonal has 'm' rows, and the lower part
// of the radical also has 'm' rows.
//
// If the user enters a value for 'm' that is not
// positive, simply return 0.

// Function declarations
int main();

// Function definitions
int main() {
    return 0;
}



























// Hint:
//
// Break the radical symbol into three sections:
//
//   1. the horizontal bar at the top,
//   2. the upper diagonal,
//   3. the lower part of the radical.
//
// Each section can be printed separately.
//
// For each row, first determine how many spaces
// must be printed before the first asterisk. Some
// rows will then require additional spaces before
// another asterisk is printed.




























// For m = 5, consider the following measurements:
//
//                        <------15----->
//    <--------20-------->***************
//             19        *
//             18       *
//             17      *
//             16     *
//             15    *
//  ^ **      12    *
//  | 2 **     9   *
//  5  4  **   6  *
//  |   6   ** 3 *
//  v    8    ***
//
// Look at each sequence separately.
//
// The horizontal bar:
//
//     spaces:      20 = 4*5
//     asterisks:   15 = 3*5
//
// The upper diagonal:
//
//     row          spaces
//      1             19
//      2             18
//      3             17
//      4             16
//      5             15
//
// The lower part:
//
//     row       leading spaces       middle spaces
//      1              0                   12
//      2              2                    9
//      3              4                    6
//      4              6                    3
//      5              8                    0
//
// Now find formulas involving 'm' and 'row'
// that generate each of these sequences.




























// For the upper diagonal, the number of leading
// spaces decreases by one on each row:
//
//     4*m - row.
//
// For the lower part, the number of leading
// spaces increases by two on each row:
//
//     2*(row - 1).
//
// The number of spaces between the two parts
// decreases by three on each row:
//
//     3*(m - row).
//
// Notice that on the final row there are no
// spaces between the two parts, so the printed
// asterisks join together.












































// Here is one solution:
/*
int main() {
    int m{};
    std::cout << "Enter a positive integer 'm': ";
    std::cin  >> m;

    if ( m <= 0 ) {
        return 0;
    }

    // Print the horizontal bar.
    for ( int column{ 1 }; column <= 4*m; ++column ) {
        std::cout << " ";
    }

    for ( int column{ 1 }; column <= 3*m; ++column ) {
        std::cout << "*";
    }

    std::cout << std::endl;

    // Print the upper diagonal.
    for ( int row{ 1 }; row <= m; ++row ) {
        for ( int column{ 1 }; column <= 4*m - row; ++column ) {
            std::cout << " ";
        }

        std::cout << "*" << std::endl;
    }

    // Print the lower part of the radical.
    for ( int row{ 1 }; row <= m; ++row ) {

        // Print the leading spaces.
        for ( int column{ 1 };
              column <= 2*(row - 1);
              ++column ) {
            std::cout << " ";
        }

        std::cout << "**";

        // Print the spaces between the two
        // parts of the current row.
        for ( int column{ 1 };
              column <= 3*(m - row);
              ++column ) {
            std::cout << " ";
        }

        std::cout << "*" << std::endl;
    }

    return 0;
}
*/