結果

問題 No.2918 Divide Applicants Fairly
ユーザー 👑 binap
提出日時 2024-10-01 14:28:14
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 898 ms / 2,000 ms
コード長 3,550 bytes
コンパイル時間 4,509 ms
コンパイル使用メモリ 259,124 KB
最終ジャッジ日時 2025-02-24 14:22:12
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 61
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}


const int INF = 1001001001;
const int K = 20;

int main(){
	int n;
	cin >> n;
	
	if(n >= 26){
		cout << "0\n";
		return 0;
	}
	
	vector<int> a(n);
	cin >> a;
	vector<int> perm(n);
	rep(i, n) perm[i] = i;
	int n1 = n / 2;
	int n2 = n - n1;
	
	vector<int> perm1(n1);
	rep(i, n1) perm1[i] = i;
	vector<int> perm2(n2);
	rep(i, n2) perm2[i] = n1 + i;
	
	auto split = [&](int n, vector<int> perm){
		vector<set<int>> res(2 * K + 1);
		rep(i, n){
			vector<set<int>> res_old(2 * K + 1);
			swap(res, res_old);
			for(int j = 1; j < 2 * K; j++){
				for(int sum : res_old[j]){
					res[j - 1].insert(sum - a[perm[i]]);
					res[j].insert(sum);
					res[j + 1].insert(sum + a[perm[i]]);
				}
			}
			{
				res[K - 1].insert(-a[perm[i]]);
				res[K + 1].insert(+a[perm[i]]);
			}
		}
		return res;
	};
	
	vector<set<int>> set1 = split(n1, perm1);
	vector<set<int>> set2 = split(n2, perm2);
	
	int ans = INF;
	for(int i = 0; i <= n1; i++){
		for(int sum1 : set1[K + i]){
			auto it = set2[K + i].lower_bound(sum1);
			if(it != set2[K + i].end()){
				chmin(ans, *it - sum1);
			}
			if(it != set2[K + i].begin()){
				chmin(ans, sum1 - *prev(it));
			}
		}
	}
	for(int sum : set1[K]) chmin(ans, abs(sum));
	for(int sum : set2[K]) chmin(ans, abs(sum));
	cout << ans << "\n";
	return 0;
}
0