結果

問題 No.2918 Divide Applicants Fairly
ユーザー 👑 binap
提出日時 2024-10-01 13:50:37
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 392 ms / 2,000 ms
コード長 3,557 bytes
コンパイル時間 4,551 ms
コンパイル使用メモリ 255,392 KB
最終ジャッジ日時 2025-02-24 14:21:33
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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);}

// thanks for https://qiita.com/mdstoy/items/39104b686540c5f1dc6c
template <typename T> bool next_combination(const T first, const T last, int k) {
    const T subset = first + k;
    // empty container | k = 0 | k == n 
    if (first == last || first == subset || last == subset) {
        return false;
    }
    T src = subset;
    while (first != src) {
        src--;
        if (*src < *(last - 1)) {
            T dest = subset;
            while (*src >= *dest) {
                dest++;
            }
            iter_swap(src, dest);
            rotate(src + 1, dest + 1, last);
            rotate(subset, subset + (last - dest) - 1, last);
            return true;
        }
    }
    // restore
    rotate(first, subset, last);
    return false;
}

const int INF = 1001001001;

int main(){
	int n;
	cin >> n;
	vector<int> a(n);
	cin >> a;
	vector<int> perm(n);
	rep(i, n) perm[i] = i;
	int m = n / 2;
	
	set<int> set_sum;
	do{
		int sum = 0;
		rep(i, m) sum += a[perm[i]];
		if(set_sum.find(sum) == set_sum.end()) set_sum.insert(sum);
		else{
			cout << "0\n";
			return 0;
		}
	}while(next_combination(perm.begin(), perm.end(), m));
	
	int prev = -INF;
	int ans = INF;
	for(int val : set_sum){
		chmin(ans, val - prev);
		prev = val;
	}
	cout << ans << "\n";
	return 0;
}
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