题意：

用双端队列来找到最长的上升子序列，并且字典序最小

分析

根据数据范围 \(1000\) 可以发现在可以\(n^2logn\),那我们就可以暴力枚举r开始的第一个点作为起点，通过上升子序列，找出来，最后求出字典序最小的即可

代码

/*made in mrd*/
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
#define int long long
#define mem(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define lu u << 1
#define ru u << 1 | 1
#define pb push_back
#define bug1(x) cout << x << endl
#define bug2(x, y) cout << x << ' ' << y << endl
#define pii pair<int, int>
#define bug3(x, y, z) cout << x << ' ' << y << ' ' << z << endl
int g1[N];
int a[N];
int f[N];
int res = 0;
int st[N];
int f2[N];
vector<int> g;
signed main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];

    for (int i = 1; i <= n + 1; i++) {
        vector<int> s;
        int len = 0;
        for (int j = i; j <= n; j++) {
            if (!len)
                f[++len] = a[j], g1[j] = len;
            else {
                if (f[len] < a[j] && a[j] > f[1])
                    f[++len] = a[j], g1[j] = len;
                else if (a[j] > f[1]) {
                    int x = lower_bound(f + 1, f + len + 1, a[j]) - f;
                    g1[j] = x;
                    f[x] = a[j];
                }
            }
        }
        int cnt = len;
        int len1 = 0;
        for (int j = n; j; j--) {
            if (g1[j] == cnt && cnt) s.pb(j), cnt--;
            g1[j]=0;
        }
        sort(s.begin(), s.end());
        for (int j = 0; j < s.size(); j++) st[s[j]] = 1;
        int x = f[1];
        if (len == 0) x = 1e18;
        for (int j = n; j ; j--) {
            if (st[j]) {
                st[j] = 0;
                continue;
            }
            if (!len1 && a[j] <x)
                f[++len1] = a[j];
            else if (a[j] < x) {
                if (a[j] > f[len1])
                    f[++len1] = a[j];
                else {
                    int x = lower_bound(f + 1, f + len1 + 1, a[j]) - f;
                    f[x] = a[j];
                }
            }
        }
        if (len + len1 > res)
            g = s, res = len + len1;
        else if (len + len1 == res) {
            if(g.size()==0) continue;
            if (s.size() == 0) {
                g = s;
            } else if (s > g)
                g = s;
        }
    }
    for (int i = 0; i < g.size(); i++) {
        st[g[i]] = 1;
    }
    cout << res << endl;
    for (int i = 1; i <= n; i++) {
        if (st[i])
            cout << "R";
        else
            cout << "L";
    }
    return 0;
}