**PROBLEM LINK:**

https://www.codechef.com/BLAS2020/problems/HLPSAM

**Contest:** Hey Newbees, Here is Some Honey(Round 2)

**Contest Link:** https://www.codechef.com/BLAS2020?itm_campaign=contest_listing

**Author:** aimon_123

**Tester:** arghya_n

**Editorialist:** aimon_123

**DIFFICULTY:**

SIMPLE.

**PREREQUISITES:**

Greedy,Math.

**Tutorial:**

In the problem , At first, you have to find how much odd number and how much even number in there from 1 to n .

Then you have to find the number of operation performed by even number denoted e in code and the number of operation performed by odd number denoted o in the code .We can prove that , total operation = number of operation using even number+number of operation using odd number =(e*(e-1))/2 + (o*(o-1))/2

Total earning of Sam is = total operation * 2 = (e*(e-1)) + (o*(o-1))

Profit / loss (denoted m in the code) = x - Total earning of Sam

If(m > 0) Print m and profit,if(m < 0) print -m and loss,if(m=0) print 0 and null.

**Source code :**

## CPP

//Code by ASIM FOIZE AIMON

//CSE’19;CUET

//avoid copy paste//Hit the friend button

#include <bits/stdc++.h>

using namespace std;

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);

#define precise cout.precision(10); cout << fixed;

#define endl “\n”

#define int int64_t

#define ll long long

#define yes printf(“YES\n”)

#define no printf(“NO\n”)

int32_t main()

{

```
ll t;
cin>>t;
while(t--)
{
ll a[100001],b[100001],c=0,d=0,e,f,m,n,p,x,y,z,i,j,k;
string s;
cin>>n>>x;
if(n%2==0)
{
c=n/2;
d=n/2;
}
else
{
c=n/2;
d=n/2+1;
}
y=(c*(c-1))/2+(d*(d-1))/2;
m=x-(y*2);
if(m<0)
cout<< -m << " " << "Profit" << endl;
else if(m>0)
cout<< m << " " << "Loss" << endl;
else
cout<< 0 << " " << " Null " << endl;
}
```

}