제출 #170885

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
170885		ne4eHbKa	Bali Sculptures (APIO15_sculpture)	C++17	100 / 100	411 ms	1016 KiB

sculpture

//{ <defines>
#ifndef _LOCAL
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#endif

#include <bits/stdc++.h>
using namespace std;

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int(x.size())
#define pw(x) (1 << (x))
#define PW(x) (1ll << (x))
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset(x, y, sizeof x)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

const int oo = 2e9;
const ll OO = 4e18;
//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int md = 0x3b800001;
const int MD = 1e9 + 7;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}
//} </defines>

void solve() {
    int n, a, b; cin >> n >> a >> b;
    ll y[n]; fo(n) cin >> y[i];

    if(n > 100) {
        const int N = 2000;
        typedef bitset<N + 1> bi;

        bi g[n + 1];
        ll ans = 0;

        for(char bt = 44; ~bt; --bt) {
            ll mx = PW(bt);

            bool fail = false;

            for(int l = 0; l < n; ++l) {
                ll sum = 0;
                for(int r = l; r < n; ++r) {
                    sum += y[r];
                    g[l][r + 1] = (sum & ~ans) < mx;
                }
            }

            bi q; q[0] = 1;
            fo(b) {
                if(q[n]) break;
                bi t;
                fo(n) if(q[i]) t |= g[i];
                q = t;
            }

            if(!q[n]) fail = true;

            if(fail) ans ^= mx;
        }

        cout << ans << endl;
    } else {
        const int N = 100;
        typedef bitset<N + 1> bi;
        bi g[n + 1];

        ll ans = OO;

        for(int x = a; x <= b; ++x) {
            ll f = 0;

            for(char bt = 38; ~bt; --bt) {
                ll mx = PW(bt);

                bool fail = false;

                for(int l = 0; l < n; ++l) {
                    ll sum = 0;
                    for(int r = l; r < n; ++r) {
                        sum += y[r];
                        g[l][r + 1] = (sum & ~f) < mx;
                    }
                }

                bi q; q[0] = 1;
                fo(x) {
                    bi t;
                    fo(n) if(q[i]) t |= g[i];
                    q = t;
                }

                if(!q[n]) fail = true;

                if(fail) f ^= mx;
            }

            umin(ans, f);
        }

        cout << ans << endl;
    }
}

/*
6 1 3
8 1 2 1 5 4
*/

int main() {
    solve();
}

• Subtask 1: 9 / 9
• Subtask 2: 16 / 16
• Subtask 3: 21 / 21
• Subtask 4: 25 / 25
• Subtask 5: 29 / 29