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 Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_macros)]
macro_rules! get {
    ([$t:ty] ; $n:expr) => {
        (0..$n)
        .map(|_| get!([$t])).collect::<Vec<_>>()
    };
    ($($t:ty),+ ; $n:expr) => {
        (0..$n).map(|_| get!($($t),+)).collect::<Vec<_>>()
    };
    ([$t:ty]) => {
        rl().split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect::<Vec<$t>>()
    };
    ($t:ty) => {
        rl().parse::<$t>().unwrap()
    };
    ($($t:ty),*) => {
        {
            let line =rl();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap()),*)
        }
    };
}
fn rl() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn main() {
    let s1 = get!(String);
    let s: Vec<char> = (" ".to_string() + &s1).chars().collect();
    let t1 = get!(String);
    let t: Vec<char> = (" ".to_string() + &t1).chars().collect();
    let mut dp = vec![vec![0; s.len() + 1]; t.len() + 1];
    //1:↖ 2:↑ 3:←
    let mut prev = vec![vec![0; s.len() + 1]; t.len() + 1];
    for j in 0..s.len() {
        for i in 0..t.len() {
            if i == 0 || j == 0 {
                continue;
            }
            if t[i] == s[j] {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                prev[i][j] = 1;
            } else if t[i] != s[j] {
                if dp[i - 1][j] < dp[i][j - 1] {
                    dp[i][j] = dp[i][j - 1];
                    prev[i][j] = 3;
                } else {
                    dp[i][j] = dp[i - 1][j];
                    prev[i][j] = 2;
                }
            }
        }
    }
    //for i in 0..dp.len() {
    //    eprintln!("{:?}", dp[i]);
    //}
    let mut si = s.len() - 1;
    let mut ti = t.len() - 1;
    let mut ans = Vec::new();
    while 1 <= si && 1 <= ti {
        if prev[ti][si] == 1 {
            ans.push(s[si]);
            si -= 1;
            ti -= 1;
        } else if prev[ti][si] == 2 {
            ti -= 1;
        } else if prev[ti][si] == 3 {
            si -= 1;
        }
    }
    ans.reverse();
    for i in ans.iter() {
        print!("{}", i);
    }
    println!("");
}