621A - Wet Shark and Odd and Even - CodeForces Solution

//Believe in yourself ,( try and try and then try ), You can do it
#include <bits/stdc++.h>
#define ll long long
#define line "\n"
using namespace std;
ll summ(vector<ll>&arr );
void fast();
/*
index = (index+1) % n; // to right or back to 0
index = (index+n-1) % n; // to left or back to n-1
*/
int main()
{
  fast();
  unsigned ll n , temp , res=0 , total =0 , smallestOdd=LLONG_MAX , sumAllOdd=0; 
  vector<ll> arr; 
  cin>>n; 
  for(ll i=0; i<n; i++){
      cin>>temp;      
      if(temp%2==0) 
        res+=temp; 
      else {
        arr.push_back(temp);  //push odds only
        smallestOdd=min(smallestOdd , temp); 
        sumAllOdd+=temp; 
        }
      total+=temp ; 
      
    }//arr is all odds
  if(total%2==0) {cout<<total; return 0; }
  ll len =arr.size(); 
//   cout<<len<<" "<<sumAllOdd<<" "<<smallestOdd<<line; 
  if(len%2==0) res+=sumAllOdd; 
  else res+=sumAllOdd-smallestOdd; 
  cout<<res; 
  //idea is even + even is even and odd + odd = even but others are gonna give me odd. 
  //so all evens are gonna make even sum so i execulde them from pushing and i push all odds only 
  //and if # of array of odds is even so all of them are part of pairs of odd + odd but if 
  // # of array of odds is odd then so all of them are part of pairs of odd + odd except
  //one which is the smallest odd cuz i wanna make max even.
  
}

void fast(){
    cin.tie(nullptr);
    cout.tie(nullptr);
    cin.sync_with_stdio(0);
}