Submission #1059038

#	Time	Username	Problem	Language	Result	Execution time	Memory
1059038		sammyuri	Bali Sculptures (APIO15_sculpture)	C++17	100 / 100	100 ms	604 KiB

sculpture

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

/*
suppose we are trying to satisfy prefix j

let dp[i][0] be the minimum number of segments required to finish at i
and dp[i][1] be the maximum
we can do transitions in O(n^2) and check if dp[n][0] or dp[n][1] is good enough
then we can build from the highest bit down
overall: O(n^2 log y)?
*/

typedef long long ll;
const ll LIMIT = (1ll << 61);

const int MAXN = 2005;
int n;
int a, b;
ll nums[MAXN];
ll dp[MAXN][2];
ll badmask;

bool check() {
    for (int i = 0; i <= n; i ++)
        dp[i][0] = 1e9, dp[i][1] = -1;
    dp[0][0] = dp[0][1] = 0;
    for (int i = 0; i < n; i ++) {
        ll cursum = 0;
        for (int j = i + 1; j <= n; j ++) {
            cursum += nums[j];
            // exceeded limit
            if (cursum >= LIMIT)
                break;
            if ((cursum & badmask) == 0) {
                dp[j][0] = min(dp[j][0], dp[i][0] + 1);
                dp[j][1] = max(dp[j][1], dp[i][1] + 1);
            }
        }
    }
    return (dp[n][0] <= b) && (dp[n][1] >= a);
}


int main() {
    cin >> n >> a >> b;
    for (int i = 1; i <= n; i ++)
        cin >> nums[i];
    ll curbit = LIMIT / 2ll;
    badmask = 0;
    ll ans = 0;
    while (curbit) {
        badmask += curbit;
        if (!check()) {
            badmask -= curbit;
            ans += curbit;
        }
        curbit /= 2;
    }
    cout << ans << endl;
}

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