#include <iostream>

// Implement a program that queries the user for an
// integer 'n' and then calculates and prints 'n!!',
// the double factorial. The double factorial appears
// in some engineering models.
//
// If n < 0, simply return 0 and do nothing.
// If 'n' is even, then n!! = n*(n - 2)*(n - 4)*...*4*2.
// If 'n' is odd,  then n!! = n*(n - 2)*(n - 4)*...*5*3*1.
//
// For example,
//     0!! =     1
//     1!! =     1
//     2!! =     2
//     3!! =     3
//     4!! =     8
//     5!! =    15
//     6!! =    48
//     7!! =   105
//     8!! =   384
//     9!! =   945
//    10!! =  3840
//    11!! = 10395
//    12!! = 46080
//
// Scroll down first for a hint and then for solutions.
//
// Note that n!! is not equal to (n!)!

// Function declarations
int main();

// Function definitions
int main() {
    int n{};
    std::cout << "Enter an integer: ";
    std::cin  >> n;

    std::cout << n << "!! = " << 1 << std::endl;

    return 0;
}
























// Hint: You could use a conditional statement
// and calculate the double factorial differently
// depending on whether 'n' is even or odd.
//
// Alternatively, observe that 0!! = 1!! = 1.
// You can initialize the result to 1, start a
// loop at 'n', and count down by 2 until there
// are no more factors to multiply.

































// This is one solution:

/*
int main() {
    int n{};
    std::cout << "Enter an integer: ";
    std::cin  >> n;

    if ( n < 0 ) {
        return 0;
    }

    int double_factorial{ 1 };

    for ( int k{ n }; k > 1; k -= 2 ) {
        double_factorial *= k;
    }

    std::cout << n << "!! = "
              << double_factorial << std::endl;

    return 0;
}
*/

// This is another solution:

/*
int main() {
    int n{};
    std::cout << "Enter an integer: ";
    std::cin  >> n;

    if ( n < 0 ) {
        return 0;
    }

    int double_factorial{ 1 };

    if ( n%2 == 0 ) {
        // 'n' is even.
        for ( int k{ 2 }; k <= n; k += 2 ) {
            double_factorial *= k;
        }
    } else {
        // 'n' is odd.
        for ( int k{ 3 }; k <= n; k += 2 ) {
            double_factorial *= k;
        }
    }

    std::cout << n << "!! = "
              << double_factorial << std::endl;

    return 0;
}
*/