結果

問題 No.2918 Divide Applicants Fairly
ユーザー 👑 binap
提出日時 2024-10-01 14:25:35
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,458 bytes
コンパイル時間 4,353 ms
コンパイル使用メモリ 258,536 KB
最終ジャッジ日時 2025-02-24 14:21:53
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 58 WA * 3
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 #
プレゼンテーションモードにする

#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));
            }
        }
    }
    cout << ans << "\n";
    return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0