Problem:
Score : 100 points

Problem Statement

You are given strings s and t.
Find one longest string that is a subsequence of both s and t.

Notes

A subsequence of a string x is the string obtained by removing zero or more characters from x and concatenating the remaining characters without changing the order.

Constraints

s and t are strings consisting of lowercase English letters.

1 \leq |s|, |t| \leq 3000

Input

Input is given from Standard Input in the following format:

s
t

Output

Print one longest string that is a subsequence of both s and t.
If there are multiple such strings, any of them will be accepted.

Sample Input 1

axyb
abyxb

Sample Output 1

axb

The answer is axb or ayb; either will be accepted.

Sample Input 2

aa
xayaz

Sample Output 2

aa

Sample Input 3

a
z

Sample Output 3

The answer is (an empty string).

Sample Input 4

abracadabra
avadakedavra

Sample Output 4

aaadara

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

# include <bits/stdc++.h>
# define sz(x) (int)((x).size())
# define int long long
# define F first
# define S second
# define pb push_back
# define all(x) x.begin(), x.end()
# define pqueue priority_queue
# define mset multiset
# define umap unordered_map
# define Speed() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
# define rep(i,n) for(int i=0;i<n;++i)

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const int N = 3005;
 
string s,t,ans;
int used[N],dp[N][N];
char ret[N];

void solve(){
	cin>>s>>t;
	s="#"+s;
	t="#"+t;
	for(int i=1;i<=sz(s);++i){
		for(int j=1;j<=sz(t);++j){
			if(s[i]==t[j]){
				dp[i][j]=dp[i-1][j-1]+1;
				if(!used[dp[i][j]]){
					used[dp[i][j]]=1;
					ret[dp[i][j]]=s[i];
				}
			}
			else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
		}
	}
	for(int i=1;i<=dp[sz(s)][sz(t)];++i){
		ans+=ret[i];
	}
	cout<<ans;
}

signed main(){
	Speed();
	//int T;cin>>T;while(T--)
	solve();
}