Hackerrank's Problem Solving solutions( Utopian Tree, Angry Professor, Beautiful Days at the Movies ) :
Problem 28: Utopian Tree
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t; //t=no. of test cases
cin>>t;
while(t--)
{
int n,h=1,i; //h=height
cin>>n;
if(n==0)
{
cout<<"1"<<endl;
}
else
{
for(i=1;i<=n;i++)
{
if(i%2==0)
{
h=h+1;
}
else
{
h=h*2;
}
}
cout<<h<<endl;
}
}
}
This code is simple. There's no need for any explanation.
Problem 29: Angry Professor
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,k,ct=0,e; //ct=count , e=arrival time
cin>>n>>k;
while(n--)
{
cin>>e;
if(e<=0)
{
ct++;
}
}
if(ct>=k)
{
cout<<"NO"<<endl;
}
else
{
cout<<"YES"<<endl;
}
}
}
This code is simple. There's no need for any explanation.
Problem 30: Beautiful Days at the Movies
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
long long rev(long long n) //function for finding the reverse of a number
{
long long reverse=0,r;
while(n!=0)
{
r=n%10;
reverse=reverse*10 + r;
n=n/10;
}
return reverse;
}
int main()
{
long long i,j,k,p,ct=0;
cin>>i>>j>>k;
for(p=i;p<=j;p++)
{
if((abs(p-rev(p))%k==0)) //doing what is asked in the question
{
ct++;
}
}
cout<<ct;
}
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