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feat: improve the Armstrong Number algorithm (#2480)
Co-authored-by: realstealthninja <68815218+realstealthninja@users.noreply.github.com>
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// Program to check whether a number is an armstrong number or not
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#include <cmath>
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#include <iostream>
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using std::cin;
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using std::cout;
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/**
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* @file
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* @brief Checks whether a number is an [Armstrong
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* Number](https://en.wikipedia.org/wiki/Narcissistic_number) or not.
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*
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* @details
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* An Armstrong number is a number that is the sum of its own digits each raised
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* to the power of the number of digits. For example: 153 is an Armstrong number
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* since 153 = 1^3 + 5^3 + 3^3.
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*
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* A few examples of valid armstrong numbers:
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* 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748,
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* 92727, 93084.
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*
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* Armstrong numbers are also known as Narcissistic Numbers, as stated in
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* Wikipedia.
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*
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* @author [Shivam Singhal](https://github.com/shivhek25)
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* @author [David Leal](https://github.com/Panquesito7)
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*/
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int main() {
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int n = 0, temp = 0, rem = 0, count = 0, sum = 0;
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cout << "Enter a number: ";
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cin >> n;
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#include <cassert> /// for assert
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#include <cmath> /// for std::pow
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#include <iostream> /// for IO operations
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temp = n;
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/**
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* @namespace
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* @brief Dynamic Programming algorithms
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*/
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namespace dynamic_programming {
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/* First Count the number of digits
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in the given number */
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/**
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* @brief Checks if the given number is armstrong or not.
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* @param number the number to check
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* @returns false if the given number is NOT armstrong
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* @returns true if the given number IS armstrong
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*/
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template <typename T>
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bool is_armstrong(const T &number) {
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int count = 0, temp = number, result = 0, rem = 0;
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// Count the number of digits of the given number.
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// For example: 153 would be 3 digits.
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while (temp != 0) {
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temp /= 10;
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count++;
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}
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/* Calculation for checking of armstrongs number i.e.
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in an n-digit number sum of the digits is raised to a power of n
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is equal to the original number */
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temp = n;
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// Calculation for checking of armstrongs number i.e.
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// in an n-digit number sum of the digits is raised to a power of `n` is
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// equal to the original number.
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temp = number;
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while (temp != 0) {
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rem = temp % 10;
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sum += static_cast<int>(pow(rem, count));
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result += static_cast<T>(std::pow(rem, count));
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temp /= 10;
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}
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if (sum == n) {
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cout << n << " is an armstrong number";
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if (result == number) {
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return true;
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} else {
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cout << n << " is not an armstrong number";
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return false;
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}
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}
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} // namespace dynamic_programming
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/**
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* @brief Self-test implementations
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* @returns void
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*/
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static void tests() {
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assert(dynamic_programming::is_armstrong(153) == true);
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assert(dynamic_programming::is_armstrong(1) == true);
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assert(dynamic_programming::is_armstrong(0) == true);
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assert(dynamic_programming::is_armstrong(370) == true);
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assert(dynamic_programming::is_armstrong(1634) == true);
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assert(dynamic_programming::is_armstrong(580) == false);
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assert(dynamic_programming::is_armstrong(15) == false);
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assert(dynamic_programming::is_armstrong(1024) == false);
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assert(dynamic_programming::is_armstrong(989) == false);
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assert(dynamic_programming::is_armstrong(103) == false);
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std::cout << "All tests have successfully passed!\n";
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}
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/**
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* @brief Main function
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* @returns 0 on exit
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*/
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int main() {
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tests(); // run self-test implementations
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return 0;
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}
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