How to resolve the algorithm Anti-primes step by step in the C++ programming language
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:
-
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 ofn
) using thefor
loop. - Checks if
n
is divisible byi
(i.e.,n%i == 0
). If it is, it increments thecount
variable becausei
is another divisor ofn
. - After the loop ends, the function returns the
count
variable, which represents the number of divisors for the given numbern
.
- Takes an integer
-
main
Function:- Initializes two integer variables:
maxDiv
to 0 (which will store the maximum number of divisors encountered) andcount
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 untilcount
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 forn
. - Checks if the number of divisors for
n
is greater thanmaxDiv
. 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.
- Prints the number
- Calls the
- The loop continues until
count
reaches 20, and the program outputs the first 20 anti-primes.
- Initializes two integer variables:
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 ofn
. - 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 initializesmaxDiv
to 0 andcount
to 0. - It enters a loop that iterates through positive integers starting from 1.
- For each
n
, it calls thecountDivisors
function to find the number of divisorsd
. - It checks if
d
is greater thanmaxDiv
. If it is, it printsn
, updatesmaxDiv
tod
, and incrementscount
. - 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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