Submission #490128

#	Time	Username	Problem	Language	Result	Execution time	Memory
490128		JerryLiu06	Bali Sculptures (APIO15_sculpture)	C++17	37 / 100	5 ms	316 KiB

sculpture

// https://oj.uz/problem/view/APIO15_sculpture

#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double;
using str = string;
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(0);
#define FASTIOF ios_base::sync_with_stdio(false); fin.tie(0);
 
#define mp make_pair
#define f first
#define s second
 
#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a)
#define EACH(a, x) for (auto& a : x)
 
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
 
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
 
template<class T> void remDup(vector<T>& v) { sor(v); v.erase(unique(all(v)), v.end()); }
 
const int MOD = 1e9 + 7;
const int MX = 2e3 + 10;
const ll INF = 1e18;

int N, A, B; vi Y; 

bool check(ll cur, int ind, ll sum) {
    FOR(i, ind + 1, 45) {
        if ((sum & (1LL << i)) && !(cur & (1LL << i))) {
            return false;
        }
    }
    return (sum & (1LL << ind)) == 0;
}

bool valid(ll cur, int ind) {
    if (A == 1) {
        int DP[N + 1];
    
        F0R(i, N + 1) {
            DP[i] = MOD;
        }
        DP[0] = 0;
    
        FOR(i, 1, N + 1) {
            ll sum = 0;
    
            R0F(j, i) {
                sum += Y[j + 1];
    
                if (check(cur, ind, sum)) {
                    ckmin(DP[i], DP[j] + 1);
                }
            }
        }
        return DP[N] <= B; 
    }
    bool DP[N + 1][B + 1];
    
    F0R(i, N + 1) {
        F0R(j, B + 1) {
            DP[i][j] = false;
        }
    }
    DP[0][0] = true;
    
    FOR(i, 1, N + 1) {
        FOR(j, 1, B + 1) {
            ll sum = 0;
            
            R0F(k, i) {
                sum += Y[k + 1];
                
                if (DP[k][j - 1]) {
                    if (check(cur, ind, sum)) {
                        DP[i][j] = true;
                        break;
                    }
                }
            }
        }
    }
    FOR(i, A, B + 1) {
        if (DP[N][i]) return true;
    }
    return false;
}

int main() {
    FASTIO;

    cin >> N >> A >> B;

    Y.rsz(N + 1); 

    FOR(i, 1, N + 1) {
        cin >> Y[i];
    }

    ll ans = 0;

    R0F(i, 45) {
        if (!valid(ans, i)) {
            ans += (1 << i);
        }
    }
    cout << ans;
}

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