# PROBLEM LINK:

Practice

Contest: Hey Newbees, Here is Some Honey

*Author:* Samia Rahman

*Tester:* Arefin Labib

*Tester:* Nazmus Sakib Rhythm

*Editorialist:* Samia Rahman

# DIFFICULTY:

SIMPLE.

# PREREQUISITES:

Math,Geometry.

# PROBLEM:

In the problem there is a given integer called N.You have to find out the Oddsum

and Evensum from 1 to N.There is also given a circle whose radius R is unknown

because you have to figure it out through N,where N is the arc length of the

circle.There is also a square given in the problem and you need to find out S ,where

S is a half diagonal of the square and N is another full diagonal of the square.Then

find out the LCM(oddsum,R) and LCM(evensum,S) and do compare them.

# QUICK EXPLANATION:

Compare L1 and L2

Where,

L1=lcm(Oddsum,R)

L2=lcm(Evensum,S)

Oddsum=ceil(N/2)xceil(N/2)

Evensum=(N/2)x((N/2)+1)

R=ceil((2*N)/pi)

S=ceil(N/2)

# EXPLANATION:

1.In case of Nova you need summation of odd numbers from 1 to N.Due to

**shortness of time limit** you need to figure this out with a formula.

Summation of odd numbers from 1 to N is-

**Oddsum=ceil(N/2)xceil(N/2).**

*To find R -*

**N is the Arc Length of circle.**

Formula of arc length-

S=rxθ

= arc length (in our code it is N)

= Radius (in our code it is R)

= central angle in radian (in our code it is θ)

So,

N=Rxθ

=Rx90 degree (WE NEED THE θ IN RADIAN)

=Rx90x(pi/180)

N=Rx(pi/2) //we all know that 90 degree=pi/2.

**R=(2xN)/pi.**

**Take the ceil value of R.**

**LCM of Nova= (OddsumxR)/__gcd(Oddsum,R);**

2.In case of Oliver Summation of odd numbers from 1 to N is-

**Evensum=(N/2)x((N/2)+1).**

*To find S-*

**S is perpendicular on the middle point of the square’s diagonal N.**

**It’s a theorem that it will be N/2.**

If you don’t know then apply Pythagoras formula-

Let’s assume side of the square is A.So N=Axsqrt(2).

So A =N/sqrt(2).

From Pythagoras formula-

A^2=(N/2)^2+S^2

Or,*S^2*=( N/sqrt(2))^2-(N/2)^2

=(N^2)/2-(N^2)/4

=(N^2)/4

*S=sqrt((N^2)/4).*

**S=N/2**.

**Take the ceil value of S.**

**LCM of Oliver= (EvensumxS)/__gcd(Evensum,S);**

**TIME COMPLEXITY: O(1)**

# SOLUTIONS:

## Setter's Solution

```
#include<bits/stdc++.h>
#define FastRead \
ios_base::sync_with_stdio(false); \
cin.tie(0);
#define ll long long
#define endl "\n"
#define pi acos(-1)
using namespace std;
int main()
{
FastRead
ll int t,j,n,e,o,m,g1,g2,l1,l2,r,s;
double p;
cin>>t;
for(j=0; j<t; j++)
{
cin>>n;
if(n%2==1)
m=(n/2)+1;
else
m=n/2;
e=(n/2)*((n/2)+1);
o=m*m;
p=(double)n;
s=ceil(p/2);
r=ceil((2*p)/pi);
g1=__gcd(e,s);
g2=__gcd(o,r);
l1=(e*s)/g1;
l2=(o*r)/g2;
if(l1>l2)
cout<<"Nova's gonna kill me"<<endl;
else
cout<<"YESS(sunglass emo)"<<endl;
}
return 0;
}
```