FASTJON Editorials

editorial
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1

PROBLEM LINK:

Contest

Setter: sayan_kashyapi

Tester: mishra_roshan

Editorialist: shaw_sandeep

DIFFICULTY:

Simple

PREREQUISITES:

FizzBuzz Program / Basic Conditional Logic

PROBLEM:

Given three integers A, B, and N. Travel from 1 to N, If the number is a multiple of A, print “Beautiful”. If the number is a multiple of B, print “Lie”. If the number is a multiple of both A and B, print “BeautifulLie” else print the current number itself.

EXPLANATION

The implementation is same as standard FizzBuzz program with some basic modification.

  1. Check for multiple of A and B i.e, “BeautifulLie”

  2. Either check for the multiple of A (Beautiful) or multiple of B (Lie)

  3. Else print the number itself

TIME COMPLEXITY

Time complexity is O(N).

SOLUTIONS:

Setter's Solution
C++ 

#include <iostream>

using namespace std;

int main(){

    int n1, n2,n;

    cin >> n1 >> n2 >> n;

    

    for ( int i = 1 ; i <= n ; i++){

        if(i%n1==0 && i%n2==0){

            cout << "BeautifulLie" << endl;

        }

        else if(i%n1==0){

            cout << "Beautiful" << endl;

        }

        else if(i%n2==0){

            cout << "Lie" << endl;

        }

        else{

            cout << i << endl;

        }

    }

    return 0;

}
Tester's Solution
C++ 

#include <iostream>

using namespace std;

int main()

{

    int n1, n2,n;

    cin >> n1 >> n2>>n;

    int i, f = n1 - 1, b = n2 - 1;

    for ( i = 1 ; i <= n ; ++i, --f, --b )

    {

        if ( f && b )cout << i;

        if ( !f ) {cout << "Beautiful"; f = n1;}

        if ( !b ) {cout << "Lie"; b = n2;}

        cout << "\n";

    }

    return 0;

}

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Suggestions are welcomed as always had been.