https://codeforces.com/contest/1157/problem/C2 

题意：给一个存在重复的元素的数组，每次可以在头或者尾取一个数，求取数最长严格递增序列的方法

题解：因为如果存在两端元素相同的情况，只能取一端，因为序列严格递增，必然不可能再取另一端的数，因此当相同时，左右分别暴力即可

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
string ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    l=1;
    r=n;
    int now=0;
    while(l<=r){
        bool u=a[l]>now;
        bool v=a[r]>now;
        //cout<<l<<" "<<r<<endl;
        if(u&&v){
            if(a[l]<a[r]){
                ans+='L';
                now=a[l++];
            }else if(a[l]>a[r]){
                ans+='R';
                now=a[r--];
            }else{
                temp=ans;
                int tempnow=now;
                for(int i=l;i<=r;i++){
                    if(a[i]>now){
                        ans+='L';
                        now=a[i];
                    }else{
                        break;
                    }
                }
                for(int i=r;i>=l;i--){
                    if(a[i]>tempnow){
                        temp+='R';
                        tempnow=a[i];
                    }else{
                        break;
                    }
                }
                if(temp.size()>ans.size())
                    swap(ans,temp);
                break;
            }
        }else if(u||v){
            if(u){
                ans+='L';
                now=a[l++];
            }else{
                ans+='R';
                now=a[r--];
            }
        }else{
            break;
        }
    }
    cout<<ans.size()<<endl;
    cout<<ans<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}