Submission #170866

#	Time	Username	Problem	Language	Result	Execution time	Memory
170866		ne4eHbKa	Bali Sculptures (APIO15_sculpture)	C++17	25 / 100	1078 ms	504 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>
const int N = 2e5 + 228;

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

    int g[n + 1][n]; clr(g, 0); int e[n + 1];
    int q[n]; int m;

    ll ans = OO;

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

        for(char bt = 38; ~bt; --bt) {
            bool fail = false;

            clr(e, 0);
            for(int l = 0; l < n; ++l) {
                ll sum = 0;
                for(int r = l; r < n; ++r) {
                    sum += y[r];
                    if((sum & ~f) < PW(bt)) g[l][e[l]++] = r + 1;
                }
            }

            m = 1; q[0] = 0;
            fo(x) {
                if(!m) { fail = true; break; }
                set<int> h;
                for(int v = 0; v < m; ++v)
                    for(int u = 0; u < e[q[v]]; ++u) h.insert(g[q[v]][u]);
                m = 0; for(int v : h) q[m++] = v;
            }

            if(!m || q[m - 1] != n) fail = true;

            if(fail) f ^= PW(bt);
        }

        umin(ans, f);
    }

    cout << ans << endl;
}

int main() {
    solve();
}

• Subtask 1: 9 / 9
• Subtask 2: 16 / 16
• Subtask 3: 0 / 21
• Subtask 4: 0 / 25
• Subtask 5: 0 / 29