Codeforces 743C. Vladik and fractions Solution

Hints:

( 1 / n ) = ( 1 /x ) + ( 1/y )

We have to form this above equation , Now place (n+1) in place of x , because if we put n instead of x then value of of y will be infinity . So , we will place the maximum value as possible So now , ( 1 / n )-( 1 /x ) = ( 1/y )

=> ( 1 / n )-( 1 / (n+1) ) = ( 1/y )

=> ( 1 / (n*(n+1) ) = ( 1 / y )

=> So , y will be the value of n*(n+1) to make form like ( 1 / n ) = ( 1 /x ) + ( 1/y )

Here , one question should be asked that why we put value ( n + 1) instead of x not other value like ( n+2 ), (n+3 )....... Because if we put (n+2) instead of x then the equation will be like in the form ( 2 / (n*(n+2) ) = ( 1 / y ) ,But this is not our expected form ...same thing for any value without only (n+1) ,

So , we can write ( 1 / (n*(n+1) ) + ( 1 / n) = ( 1 / n )

( 2 / n ) = ( 1 / n ) + ( 1/ n)

=> ( 1 / n ) + ( 1 / (n*(n+1) ) + ( 1 / n)

So , now ( 2/n ) can be expressed as sum of three expected fractions . Be happy and implement now

Solution in C++:

///La ilaha illellahu muhammadur rasulullah

///******Bismillahir-Rahmanir-Rahim******///

///Abul Hasnat  Tonmoy

///Department of CSE,23rd batch

///Islamic University,Bangladesh

#include<iostream>

using namespace std;

int main()

{

    long long n;

    cin>>n;

    if(n==1)

    cout<<"-1"<<endl;

    else

    cout<<n<<" "<<n+1<<" "<<n*(n+1)<<endl;

}