絕世好題 BZOJ - 4300

題目傳送門

思路：這個題如果用O(N^2)的算法的話會超時，所以我們可以對每一個數字的每一位數字進行操作，這樣可以在O(N*logN)的時間複雜度中完成。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define MAXN 100010
#define MAXE 40
#define INF 0x7ffffff
#define MOD 100003
#define LL long long
#define ULL unsigned long long
#define pi 3.14159

using namespace std;

LL arr[MAXN];
int f[MAXE];

int main() {
    std::ios::sync_with_stdio(false);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }
    int max_length = 0;
    memset(f, 0, sizeof(f));
    for (LL i = 0; i < n; ++i) {
        int sum = 0;
        for (LL j = 0; j <= 30; ++j) {
            if (arr[i] & (1LL << j)) {
                sum = max(sum, f[j] + 1);
            }
        }
        for (int j = 0; j <= 30; ++j) {
            if (arr[i] & (1LL << j)) {
                f[j] = max(sum, f[j]);
            }
        }
    }
    for (int i = 0; i <= 30; ++i)
        max_length = max(max_length, f[i]);
    cout << max_length << endl;
    return 0;
}