 # HLPSAM-Editorial

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

Contest: Hey Newbees, Here is Some Honey(Round 2)
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,b,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;
}
``````

}

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