README.md

This is solution for a problem from the Recourse Center's application: https://www.recurse.com/apply

Write a program that prints out the numbers 1 to 100 (inclusive). If the number is divisible by 3, print Crackle instead of the number. If it's divisible by 5, print Pop. If it's divisible by both 3 and 5, print CracklePop. You can use any language.

CracklePop.c

#include <iostream>

using namespace std;

void main()
{
    for (int i = 1; i <= 100; i++)
    {
        bool divisible = false;
        if (i % 3 == 0)
        {
            cout << "Crackle";
            divisible = true;
        }
        if (i % 5 == 0)
        {
            cout << "Pop";
            divisible = true;
        }
        if (!divisible) cout << i;
        cout << "\n";
    }
}

CracklePop.c.min.159.c

#include <iostream>
#define P(s) std::cout<<s
void main(){for(int i=1;i<=100;i++){int d;if(d=i%3==0)P("Crackle");if(i%5==0){P("Pop");d=1;}if(!d)P(i);P("\n");}}

CracklePop.py

from __future__ import print_function #future proofing

def isDivisible(number, divisor):
    return number % divisor == 0 # check if the residual from division is zero

for i in range(1,101):
    output = ""
    if isDivisible(i, 3):
        output += "Crackle"
    if isDivisible(i, 5):
        output += "Pop"
    print(i if output == "" else output)
    

CracklePop.py.min.105.py

for i in range(1,101):
  o="Crackle" if i%3==0 else ""+"Pop" if i%5==0 else ""
  print(i if o=="" else o)