제출 #407014

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
407014		tranxuanbach	Bali Sculptures (APIO15_sculpture)	C++17	71 / 100	88 ms	580 KiB

sculpture

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define endl '\n'
#define fi first
#define se second
#define For(i, l, r) for (int i = l; i < r; i++)
#define ForE(i, l, r) for (int i = l; i <= r; i++)
#define FordE(i, l, r) for (int i = l; i >= r; i--)
#define Fora(v, a) for (auto v: a)
#define bend(a) a.begin(), a.end()
#define isz(a) ((signed)a.size())

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <pii> vpii;
typedef vector <vi> vvi;

const int N = 2e3 + 5;

int n, L, R;
ll a[N];

ll ans;
bool dp1[N][N];
int dp2[N];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("KEK.inp", "r", stdin);
    // freopen("KEK.out", "w", stdout);
    cin >> n >> L >> R;
    ForE(i, 1, n){
        cin >> a[i]; a[i] += a[i - 1];
    }
    if (n <= 100){
        FordE(bit, 36, 0){
            ll tans = ans + (1ll << bit) - 1;
            ForE(i, 1, n){
                ForE(j, 1, R){
                    dp1[i][j] = 0;
                }
            }
            dp1[0][0] = 1;
            ForE(i, 1, n){
                ForE(j, 1, R){
                    ForE(ti, 1, i){
                        ll sum = a[i] - a[ti - 1];
                        if ((tans & sum) == sum){
                            dp1[i][j] |= dp1[ti - 1][j - 1];
                        }
                    }
                }
            }
            bool ck = 0;
            ForE(j, L, R){
                ck |= dp1[n][j];
            }
            if (!ck){
                ans |= (1ll << bit);
            }
        }
        cout << ans << endl;
        return 0;
    }
    assert(L == 1);
    FordE(bit, 36, 0){
        ll tans = ans + (1ll << bit) - 1;
        ForE(i, 1, n){
            dp2[i] = N;
        }
        dp2[0] = 0;
        ForE(i, 1, n){
            ForE(ti, 1, i){
                ll sum = a[i] - a[ti - 1];
                if ((tans & sum) == sum){
                    dp2[i] = min(dp2[i], dp2[ti - 1] + 1);
                }
            }
        }
        if (dp2[n] > R){
            ans |= (1ll << bit);
        }
    }
    cout << ans << endl;
}

/*
n so, phan thanh x (a <= x <= b) khoang sao cho or cua tong la be nhat
nhieu kha nang sub 4 va sub 5 giai khac nhau vi sub 5 co a = 1 ????
xet bit cao nhat xong roi kiem tra xem co dat duoc tong do hay ko?
dp[i][j] = chekc xem phan duoc thanh j doan co or la submask cua X hay ko
update n^2?
n^3 * 30 la xong sub 1-4
a = 1 thi
xet xem so doan duoc phan it nhat la bao nhieu a
==================================================+
INPUT:                                            |
--------------------------------------------------|
6 1 3
8 1 2 1 5 4
--------------------------------------------------|
==================================================+
OUTPUT:                                           |
--------------------------------------------------|
11
--------------------------------------------------|
==================================================+
*/

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