Codesforces 467E Alex and Complicated Task

E. Alex and Complicated Task

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After you have read all the problems, probably, you think Alex is genius person. That's true! One day he came up with the following task.

Given a sequence of integer numbers a1, a2, ..., an. You are to find a longest sequence b1, b2, ..., b4m, that satisfies the following conditions:

• b4k + 1 = b4k + 3 for all valid integer k;
• b4k + 2 = b4k + 4 for all valid integer k;
• sequence b is subsequence of a (not necessarily contiguous subsequence).

And finally... Alex had given this complicated task to George, and George gave it to you. Help George to cope with the task.

Input

The first line contains a single integer n (1 ≤ n ≤ 5·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In the first line print a single integer 4m — the maximal possible length of required sequence b. In the second line print 4m integersb1, b2, ..., b4m, that is required sequence.

If there are multiple optimal answers you may print any of them.

Sample test(s)

input

4
3 5 3 5

output

4
3 5 3 5

input

10
35 1 2 1 2 35 100 200 100 200

output

8
1 2 1 2 100 200 100 200

题目需要求一个最长的子序列 ， 序列符合周期为4 ，然后奇数位，偶数位分别相等。

代码参考自美国coder : ecnerwal

#include<bits/stdc++.h>

using namespace std;

vector<int> res;
multiset<int> occ; //count number of occurrences
map<int, int> wrap; // wrapper for each value
vector<int> vals; // unwrapped values

void finish(int a, int b) {
    res.push_back(a);
    res.push_back(b);
    res.push_back(a);
    res.push_back(b);
    occ.clear();
    wrap.clear();
    vals.clear();
}

int main() {
    int N;
    cin >> N;
    for(int i = 0; i < N; i++) {
        int v; cin >> v;
        if(wrap.count(v)) {
            finish(wrap[v], v);
            continue;
        }
        else {
            if(occ.count(v)) {
                int cnt = (occ.count(v) >= 2) ? 1 : 0;
                while(cnt || vals.back() != v) {
                    if(vals.back() == v) cnt--;
                    wrap[vals.back()] = v;
                    vals.pop_back();
                }
            }
            occ.insert(v);
            vals.push_back(v);
        }
    }
    cout << res.size() << endl;
    for(int i = 0  ; i < res.size(); ++i ) cout << res[i] << ' '; cout << endl;
}