Submission #1195004


Source Code Expand

#include <iostream>
#include <string>
#include <map>
#include <algorithm>
#include <list>
#include <vector>
#include <set>
#include <stdio.h>
#include <queue>
#include <stack>
#include <deque>
#include <math.h>
#include <sstream>
#include <stdlib.h>
#include <functional>
using namespace std;
 
#define rep(i,n) for(int i = 0; i < n; i++)
#define rrep(i,a,n) for(int i = a; i < n; i++)
#define INF (1<<29)
#define INFL 0x3f3f3f3f3f3f3f3fLL
#define MOD 1000000007
#define fi first
#define se second 
#define pb push_back
#define PI 3.14159265358979323846
#define all(o) (o).begin(), (o).end()
#define rall(x) x.rbegin(),x.rend()
 
 
typedef double ld;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<ld> vd;
typedef vector < vc > vvc;
typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long int ulli;
 
const int dx[] = { 1, 0, -1, 0 };
const int dy[] = { 0, 1, 0, -1 };
 
template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
 
const ld eps = 1e-10, pi = acos(-1.0);
 
#define MAX_N 100001
 
// combination
 
ll fact[MAX_N], factinv[MAX_N];
ll mod_pow(ll n, ll p, ll m) {
	ll a = n;
	ll x = 1;
	while (p) {
		if (p & 1) x = (x * a) % m;
		a = (a * a) % m;
		p >>= 1;
	}
	return x;
}
int extgcd(int a, int b, int& x, int&y) {
	int d = a;
	if (b != 0) {
		d = extgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}
int mod_inverse(int a, int m) {
	int x, y;
	extgcd(a, m, x, y);
	return (m + x % m) % m;
}
ll inv(ll n) {
	return mod_pow(n, MOD - 2, MOD);
}
void combInit() {
	fact[0] = 1;
	rrep(i, 1, MAX_N) fact[i] = fact[i - 1] * i % MOD;
	factinv[MAX_N - 1] = inv(fact[MAX_N - 1]);
	for (int i = MAX_N - 2; i >= 0; i--) factinv[i] = factinv[i + 1] * (i + 1) % MOD;
}
 
// ダイクストラ
/*
struct edge { int to, cost; };
vector < vector < edge > > G;
vi d;
void dijkstra(int s) {priority_queue<pii, vector< pii >, greater<pii> > que;d[s] = 0;que.push(pii(0, s));while (!que.empty()) {pii p = que.top(); que.pop();int v = p.second;if (d[v] < p.first) continue;rep(i, G[v].size()) {edge e = G[v][i];int cost = e.cost;if (d[v] + cost < d[e.to]) {d[e.to] = d[v] + e.cost;que.push(pii(d[e.to], e.to));}}}}
*/
 
// Union-Find木
int par[MAX_N], rnk[MAX_N], unionSize[MAX_N];
// はじめは全ての頂点が根
void UnionFindTreeInit(int n) {
	rep(i, n) {
		par[i] = i;
		rnk[i] = 0;
		unionSize[i] = 1;
	}
}
//木の根元を求める
int root(int x) {
	if (par[x] == x) return x; // 根を返す
	else return par[x] = root(par[x]);
}
// 連結成分の大きさを取得
int componentSize(int x) {
	return unionSize[root(x)];
}
// xとyが同じ集合に属するか否か
bool same(int x, int y) {
	return root(x) == root(y);
}
// xとyの属する集合を併合
void unite(int x, int y) {
	x = root(x);
	y = root(y);
	if (x == y) return;
	if (rnk[x] < rnk[y]) {
		par[x] = y;
		unionSize[y] += unionSize[x];
	}
	else {
		par[y] = x;
		if (rnk[x] == rnk[y]) rnk[x]++;
		unionSize[x] += unionSize[y];
	}
}
 
 
//-----------------------------------------------------------------
 
//int dx[8] = { -1, 1, 0, 0, 1, 1, -1, -1 };
//int dy[8] = { 0, 0, 1, -1, 1, -1, -1, 1 };
 
struct edge { int to, cost; };
vector < vector < edge > > G;
vi d;
void dijkstra(int s) {
	priority_queue<pii, vector< pii >, greater<pii> > que;
	d[s] = 0;
	que.push(pii(0, s));
 
	while (!que.empty()) {
		pii p = que.top();
		que.pop();
		int v = p.second;
		if (d[v] < p.first) continue;
 
		rep(i, G[v].size()) {
			edge e = G[v][i];
			int cost = e.cost;
			if (d[v] + cost < d[e.to]) {
				d[e.to] = d[v] + e.cost;
				que.push(pii(d[e.to], e.to));
			}
		}
	}
 
}
 
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
    ll n;
    cin >> n;
    vector<ll> a(n * 3);
    rep(i,n*3) cin >> a[i];
    sort(all(a));
    ll ans = 0;
    rep(i, n) {
        ans += a[i * 2 + n];
    }
    cout << ans << endl;


	return 0;
 
}
 
 
 
 

Submission Info

Submission Time
Task A - AtCoder Group Contest
User youki
Language C++14 (GCC 5.4.1)
Score 300
Code Size 4222 Byte
Status AC
Exec Time 46 ms
Memory 2560 KB

Judge Result

Set Name Sample All
Score / Max Score 0 / 0 300 / 300
Status
AC × 2
AC × 12
Set Name Test Cases
Sample 00_example_01.txt, 00_example_02.txt
All 00_example_01.txt, 00_example_02.txt, 01.txt, 02.txt, 03.txt, 04.txt, 05.txt, 06.txt, 07.txt, 08.txt, 09.txt, 10.txt
Case Name Status Exec Time Memory
00_example_01.txt AC 1 ms 256 KB
00_example_02.txt AC 1 ms 256 KB
01.txt AC 1 ms 256 KB
02.txt AC 1 ms 256 KB
03.txt AC 1 ms 256 KB
04.txt AC 1 ms 256 KB
05.txt AC 4 ms 384 KB
06.txt AC 36 ms 2560 KB
07.txt AC 22 ms 2560 KB
08.txt AC 46 ms 2560 KB
09.txt AC 45 ms 2560 KB
10.txt AC 45 ms 2560 KB