Submission #1261676

#	Time	Username	Problem	Language	Result	Execution time	Memory
1261676		euclid	Longest beautiful sequence (IZhO17_subsequence)	C++20	100 / 100	3648 ms	187996 KiB

subsequence

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
#define pb push_back
#define fi first
#define se second
#define vi vector<int>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vl vector<ll>

int n;
const int MN = 1e5 + 7, MS = (1<<20)+7, MB = (1<<10)+7, MK = 21;
int a[MN], k[MN], best[MN], prv[MN];
int dp[MB][MB][MK][2]; //dp[l][r][k][0]: for all x with l(x)=l, bitcount(r(x), r) = k,
//the maximum length of an LBS ending at x, dp[l][r][k][1] is the index of this optimal x
int bc[MB][MB]; //bit count of (i&j)
int l[MS], r[MS]; //l[i]: leftmost 10 bits of i, r[i]: rightmost 10 bits of i

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    for(int i = 0; i < MB; i++)
        for(int j = 0; j < MB; j++) bc[i][j] = __builtin_popcount(i&j);
    for(int i = 0; i < MS; i++) {
        l[i] = i>>10;
        r[i] = i % (1<<10);
    }
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i <= n; i++) cin >> k[i];
    for(int i = 1; i <= n; i++) {
        for(int lx = 0; lx < (1<<10); lx++) {
            if(k[i] < bc[lx][l[a[i]]]) continue;
//            if(i == 1 && lx == 2) cout << best[i] << ' ' << r[a[i]] << ' ' << bc[lx][l[a[i]]] << ' ' << dp[lx][r[a[i]]][k[i]-bc[lx][l[a[i]]]][0] << '\n';
            int cur = dp[lx][r[a[i]]][k[i]-bc[lx][l[a[i]]]][0];
            if(cur+1 > best[i]) {
                best[i] = cur+1;
                prv[i] = dp[lx][r[a[i]]][k[i]-bc[lx][l[a[i]]]][1];
            }
        }
        for(int y = 0; y < (1<<10); y++) {
            if(best[i] > dp[l[a[i]]][y][bc[y][r[a[i]]]][0]) {
                dp[l[a[i]]][y][bc[y][r[a[i]]]][0] = best[i];
                dp[l[a[i]]][y][bc[y][r[a[i]]]][1] = i;
            }
        }
    }
    int mxidx = 1;
    for(int i = 1; i <= n; i++) {
//        cout << best[i] << ' ';
        if(best[i] > best[mxidx]) mxidx = i;
    }
    cout << best[mxidx] << '\n';
    vi ans;
    while(mxidx != 0) {
        ans.pb(mxidx);
        mxidx = prv[mxidx];
    }
    reverse(ans.begin(), ans.end());
    for(int num : ans) cout << num << ' ';

//    4
//    1 2 3 4
//    10 0 1 0
//    5
//    5 3 5 3 5
//    10 1 20 1 20

}

• Subtask 1: 7 / 7
• Subtask 2: 16 / 16
• Subtask 3: 17 / 17
• Subtask 4: 60 / 60