Submission #244140

#	Time	Username	Problem	Language	Result	Execution time	Memory
244140		AmineWeslati	Best Place (NOI17_bestplace)	C++14	51 / 100	40 ms	2556 KiB

bestplace

//Never stop trying
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
//s.order_of_key(), *s.find_by_order()

using namespace std;
#define boost ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

typedef string str;
typedef long long ll;
#define int ll
typedef double db;
typedef long double ld;

typedef pair<int,int> pi;
#define fi first
#define se second

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<pi> vpi;

#define pb push_back
#define eb emplace_back
#define pf push_front

#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define endl "\n"

const int MOD = 1e9+7; //998244353
const ll INF = 1e18;
const int nx[4]={0,0,1,-1}, ny[4]={1,-1,0,0}; //right left down up

int32_t main(){
   boost;
   int N; cin>>N;
   if(N==2){
      ld s=0,ss=0;
      for(int i=0; i<N; i++){
      int x,y; cin>>x>>y;
      s+=x;
      ss+=y;
      }
      cout << llround(s/N) << ' ' << llround(ss/N) << endl;
   }
   /*else if(N<=100){

   }*/
   else{
      vi vec;
      for(int i=0; i<N;i++){
         int x,y; cin>>x>>y;
         vec.pb(x);
      }
      sort(all(vec));
      /*int pref[N],suf[N];
      for(int i=0; i<N; i++){
         pref[i]=0;
         if(i) pref[i]=vec[i]-vec[i-1]+pref[i-1];

      }
      for(int i=N-1; i>=0; i--){
         suf[i]=0;
         if(i!=N-1)
            suf[i]=suf[i+1]+vec[i+1]-vec[i];
      }
      int res=INF,x,y;
      for(int i=0; i<N; i++) if(pref[i]+suf[i]<res){
         res=pref[i]+suf[i];
         x=vec[i],y=0;
      }
      cout << x << ' ' << y << endl;*/
      cout << vec[(N+1)/2-1] << ' ' << 0 << endl;
   }


   return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 20 / 20
• Subtask 3: 28 / 28
• Subtask 4: 0 / 13
• Subtask 5: 0 / 17
• Subtask 6: 0 / 19