Increasing Subsequence (hard version)

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

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