How to resolve the algorithm Anti-primes step by step in the C++ programming language

Published on 7 June 2024 03:52 AM

How to resolve the algorithm Anti-primes step by step in the C++ programming language

Table of Contents

Problem Statement

The anti-primes (or highly composite numbers, sequence A002182 in the OEIS) are the natural numbers with more factors than any smaller than itself.

Generate and show here, the first twenty anti-primes.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Anti-primes step by step in the C++ programming language

This C++ program finds the first 20 anti-primes, which are numbers with the most divisors.

Here's a step-by-step explanation of the code:

  1. countDivisors Function:

    • Takes an integer n as input.
    • If n is less than 2, it returns 1 because all numbers less than 2 have only one divisor (themselves), and 1 is not considered an anti-prime.
    • Initializes a variable count to 2 (because all numbers have at least 1 and themselves as divisors).
    • Loops through all numbers from 2 to n/2 (up to the square root of n) using the for loop.
    • Checks if n is divisible by i (i.e., n%i == 0). If it is, it increments the count variable because i is another divisor of n.
    • After the loop ends, the function returns the count variable, which represents the number of divisors for the given number n.
  2. main Function:

    • Initializes two integer variables: maxDiv to 0 (which will store the maximum number of divisors encountered) and count to 0 (which will keep track of the number of anti-primes found).
    • Prints the header "The first 20 anti-primes are:" to display the result.
    • Enters a for loop that will iterate through numbers starting from 1 until count reaches 20 (i.e., finding the first 20 anti-primes).
    • For each n in the loop:
      • Calls the countDivisors function to determine the number of divisors for n.
      • Checks if the number of divisors for n is greater than maxDiv. If it is:
        • Prints the number n to the console.
        • Updates maxDiv with the new maximum number of divisors.
        • Increments count because an anti-prime with a higher number of divisors has been found.
    • The loop continues until count reaches 20, and the program outputs the first 20 anti-primes.

In summary, this program uses a loop to calculate the number of divisors for each integer starting from 1. It keeps track of the maximum number of divisors encountered and identifies the first 20 numbers with the most divisors, which are known as anti-primes.

Code Overview:

This C++ code finds and prints the first 20 anti-primes, which are positive integers with more divisors than any smaller positive integer.

Function countDivisors:

  • This function takes an integer n as input and returns the number of positive divisors of n.
  • It initializes the count to 1 (for the divisor 1) and adds 1 to the count for each divisor from 2 up to n/2.
  • If n is less than 2, it means it has only 1 divisor, so it returns 1.

Main Function:

  • The main function initializes maxDiv to 0 and count to 0.
  • It enters a loop that iterates through positive integers starting from 1.
  • For each n, it calls the countDivisors function to find the number of divisors d.
  • It checks if d is greater than maxDiv. If it is, it prints n, updates maxDiv to d, and increments count.
  • The loop continues until count reaches 20.

Algorithm:

The algorithm works by finding anti-primes in ascending order of the number of divisors. For each n, it checks if it has more divisors than any previously encountered integer. If so, it prints n and updates the maximum number of divisors seen so far. The first 20 such numbers are the anti-primes.

Output:

The output of the program will be a list of 20 anti-primes, with each anti-prime having more divisors than the previous one. For example:

The first 20 anti-primes are:
1 2 4 6 12 24 36 48 60 120 180 240 360 720 840 1260 1680 2520 5040 7560

Source code in the cpp programming language

#include <iostream>

int countDivisors(int n) {
    if (n < 2) return 1;
    int count = 2; // 1 and n
    for (int i = 2; i <= n/2; ++i) {
        if (n%i == 0) ++count;
    }
    return count;
}

int main() {
    int maxDiv = 0, count = 0;
    std::cout << "The first 20 anti-primes are:" << std::endl;
    for (int n = 1; count < 20; ++n) {
        int d = countDivisors(n);
        if (d > maxDiv) {
            std::cout << n << " ";
            maxDiv = d;
            count++;
        }
    }
    std::cout << std::endl;
    return 0;
}


  

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