| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 133859 | Kastanda | Longest beautiful sequence (IZhO17_subsequence) | C++11 | 0 ms | 0 KiB | 
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
// ItnoE
#include<bits/stdc++.h>
using namespace std;
const int N = 100005, SQ = 1024;
int n, A[N], P[N];
pair < int , int > dp[21][SQ][SQ];
inline void Max(pair < int , int > &a, pair < int , int > b) {a = max(a, b);}
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++)
        scanf("%d", &A[i]);
    int Mx = 0, Bi = 0;
    for (int i = 1; i <= n; i ++)
    {
        int k, a = A[i];
        scanf("%d", &k);
        pair < int , int > Bst = {0, 0};
        for (int j = 0; j < SQ; j ++)
            if (k >= __builtin_popcount(a >> 10 & j))
                Max(Bst, dp[k - __builtin_popcount(a >> 10 & j)][a & 1023][j]);
        P[i] = Bst.second; Bst.first ++; Bst.second = i;
        for (int j = 0; j < SQ; j ++)
            Max(dp[__builtin_popcount(a & j)][j][a >> 10], Bst);
        if (Bst.first > Mx)
            Mx = Bst.first, Bi = i;
    }
    vector < int > R;
    while (Bi)
        R.push_back(Bi), Bi = P[Bi];
    reverse(R.begin(), R.end()
    printf("%d\n", (int)R.size());
    for (int v : R)
        printf("%d ", v);
    return 0;
}
