#include <iostream>

// Write a program that queries the user for an
// integer 'n' and then prints n!.
//
// If n < 0, simply return 0. Otherwise,
//
//     0! = 1! = 1
//
// and, for n >= 2,
//
//     n! = n*(n - 1)*(n - 2)*...*3*2*1.
//
// For example,
//
//     5! = 5*4*3*2*1 = 120.
//
// Scroll down for a hint first, and then for
// two possible solutions.

// 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:
//
// Create a local variable that stores the
// value of the product. Then multiply that
// variable by each integer from 1 through 'n'.
//
// What should the initial value of that local
// variable be?
//
// Recall that when accumulating a sum, we
// normally begin with 0. When accumulating a
// product, what value should we begin with?
//
// Also consider what happens when n == 0.
// The loop should execute zero times, but the
// result must still be 1.

































// Here is one solution, multiplying upward
// from 1 through 'n':

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

    if ( n < 0 ) {
        return 0;
    }

    int factorial{ 1 };

    for ( int k{ 1 }; k <= n; ++k ) {
        factorial *= k;
    }

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

    return 0;
}
*/

// Here is another solution, multiplying downward
// from 'n' through 1:

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

    if ( n < 0 ) {
        return 0;
    }

    int factorial{ 1 };

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

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

    return 0;
}
*/