Go: package main

import (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

const (
	initialBufSize = 100000
	maxBufSize     = 1000000
)

var sc *bufio.Scanner

func initScanner(r io.Reader) *bufio.Scanner {
	buf := make([]byte, initialBufSize)

	sc := bufio.NewScanner(r)
	sc.Buffer(buf, maxBufSize)
	sc.Split(bufio.ScanWords) // bufio.ScanLines
	return sc
}

func main() {
	sc = initScanner(os.Stdin)
	resolve()
}

func resolve() {
	n := scanInt(sc)
	results := []int{}
	for i := 0; i < n; i++ {
		results = append(results, resolveInstance())
	}
	for _, v := range results {
		fmt.Println(v)
	}
	return
}

func resolveInstance() int {
	x, y := scanString(sc), scanString(sc)
	lx, ly := len(x), len(y)
	dp := make([][]int, lx+1)
	dp[0] = make([]int, ly+1)
	for i := 1; i <= lx; i++ {
		dp[i] = make([]int, ly+1)
		dp[i][0] = dp[i-1][0]
		for j := 1; j <= ly; j++ {
			dp[i][j] = dp[i-1][j]
			if dp[i][j] < dp[i][j-1] {
				dp[i][j] = dp[i][j-1]
			}
			if x[i-1] == y[j-1] {
				cL := dp[i-1][j-1] + 1
				if dp[i][j] < cL {
					dp[i][j] = cL
				}
			}
		}
	}
	return dp[lx][ly]
}

const (
	remove = 0 + iota
	match
)

// snip-scan-funcs
func scanInt(sc *bufio.Scanner) int {
	sc.Scan()
	i, _ := strconv.Atoi(sc.Text())
	return int(i)
}
func scanString(sc *bufio.Scanner) string {
	sc.Scan()
	return sc.Text()
}

C++: #define _GLIBCXX_DEBUG
#include "bits/stdc++.h"
//#include <intrin.h>  //AtCoder (gcc) 上ではこれがあると動かない。__popcnt用のincludeファイル。
using namespace std;

typedef long long ll;
typedef long double ld;


#define int long long
#define rep(i, n) for(long long i = 0; i < (n); i++)
#define sqrt(d) pow((long double) (d), 0.50)

const int INF = 2000000000; //2e9
const int large_P = 1000000007; //1e9 + 7



//繰り返し2乗法
//N^aの、Mで割った余りを求める。
ll my_pow(ll N, ll a, ll M) {
	ll tempo;
	if (a == 0) {
		return 1;
	}
	else {
		if (a % 2 == 0) {
			tempo = my_pow(N, a / 2, M);
			return (tempo * tempo) % M;
		}
		else {
			tempo = my_pow(N, a - 1, M);
			return (tempo * N) % M;
		}
	}
}



//N_C_a を M で割った余り
ll my_combination(ll N, ll a, ll M) {
	ll answer = 1;

	rep(i, a) {
		answer *= (N - i);
		answer %= M;
	}

	rep(i, a) {
		answer *= my_pow(i + 1, M - 2, M);
		answer %= M;
	}

	return answer;
}



ll my_gcd(ll& a, ll& b) {
	if (b == 0) return a;
	ll temp = a % b;
	return my_gcd(b, temp);
}



// base を底としたときの、n の i桁目を、v.at(i) に入れる。(桁数は n に応じて自動で設定する。)
void ll_to_vector(int base, long long n, vector<int>& v) {
	long long tempo = n;
	long long tempo2 = n;

	//ll n_digit = floor(log10(n) / log10(base)) + 1;  // 桁の計算を1行で。ただ誤差があるのでよくない。
	int n_digit = 1;
	while (tempo2 >= 10) {
		tempo2 /= 10;
		n_digit++;
	}

	v.assign(n_digit, 0);

	//ll n_digit = v.size();  // v のサイズをそのままにする場合。

	for (int i = 0; i < n_digit; i++) {
		v.at(i) = tempo / pow(base, n_digit - 1 - i);
		tempo -= v.at(i) * pow(base, n_digit - 1 - i);
	}
}



int char_to_int(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2; case '3': return 3; case '4': return 4;
	case '5': return 5; case '6': return 6; case '7': return 7; case '8': return 8; case '9': return 9;
	default: return 0;
	}
}


//エラトステネスの篩で、prime で ないところに false を入れる。
//true で初期化された vector を代入する必要がある。
void prime_judge(vector<bool>& prime_or_not) {

	prime_or_not.at(0) = false;
	prime_or_not.at(1) = false;

	ll n = prime_or_not.size() - 1;
	ll region = (ll)sqrt(n) + 1;

	for (ll i = 2; i <= n / 2; i++) {
		prime_or_not.at(2 * i) = false;
	}


	for (ll i2 = 1; i2 < n / 2; i2++) {
		ll i = 2 * i2 + 1;  //ここからは奇数のみ探索
		if (i < region && prime_or_not.at(i)) {
			ll j = i * i;
			while (j < n + 1) {
				prime_or_not.at(j) = false;
				j += 2 * i;
			}
		}
	}
};



map<ll, ll> divide_to_prime(int target) {
	map<ll, ll> res;

	//sqrt(target) まで調べる。
	ll upper_lim = ceil(sqrt(target));
	vector<bool> prime_or_not(upper_lim + 1, true);

	if (upper_lim < 20) prime_or_not.assign(25, true);
	prime_judge(prime_or_not);

	ll tempo = target;
	rep(i, (upper_lim + 1)) {
		if (prime_or_not.at(i)) {
			while (tempo % i == 0) {
				tempo /= i;
				res[i]++;
			}
		}
	}

	if (tempo != 1) res[tempo]++; //sqrt(target) より大きな素因数は高々1つしかない。



	return res;
}


const int N_max = 1000;


//第一変数 i; Xを何文字目まで見たか。(1 ~)
//第二変数 j; Yを何文字目まで見たか。
vector<int> dp[N_max + 1][N_max + 1];
//長さ, Xの使っている最後の位置、Yの最後の位置

signed main() {
	int q; cin >> q;



	rep(i, q) {
		string X, Y;
		cin >> X >> Y;

		int Nx = X.size();
		int Ny = Y.size();

		for (int j = 0; j <= Nx; j++) {
			dp[j][0].push_back(0);
			dp[j][0].push_back(0);
			dp[j][0].push_back(0);
		}
		for (int j = 0; j <= Ny; j++) {
			dp[0][j].push_back(0);
			dp[0][j].push_back(0);
			dp[0][j].push_back(0);
		}

		for (int j = 1; j <= Nx; j++) {
			for (int k = 1; k <= Ny; k++) {
				dp[j][k].assign(3, 0);
				dp[j][k].at(0) = max(dp[j - 1][k].at(0), dp[j][k - 1].at(0));
				
				if (dp[j][k].at(0) == dp[j - 1][k].at(0)) {
					dp[j][k].at(1) = dp[j - 1][k].at(1);
					dp[j][k].at(2) = dp[j - 1][k].at(2);
				}
				else {
					dp[j][k].at(1) = dp[j][k - 1].at(1);
					dp[j][k].at(2) = dp[j][k - 1].at(2);
				}


				if (X.at(j - 1) == Y.at(k - 1)) {
					if (dp[j][k].at(0) == dp[j - 1][k].at(0) && dp[j - 1][k].at(2) != k) {
						dp[j][k].at(0)++;
						dp[j][k].at(1) = j;
						dp[j][k].at(2) = k;
					}
					else if (dp[j][k].at(0) == dp[j][k - 1].at(0) && dp[j][k - 1].at(1) != j){
						dp[j][k].at(0)++;
						dp[j][k].at(1) = j;
						dp[j][k].at(2) = k;
					}
					
					
				}
			}
		}


		cout << dp[Nx][Ny].at(0) << endl;

		rep(j, Nx + 1) {
			rep(k, Ny + 1) {
				//cout << dp[j][k].at(0) << " ";
			}
			//cout << endl;
		}
	}

}

C++: #include <bits/stdc++.h>
using namespace std;
using ll=long long;
using vb=vector<bool>;
using vvb=vector<vb>;
using vd=vector<double>;
using vvd=vector<vd>;
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using pll=pair<ll,ll>;
using tll=tuple<ll,ll>;
using vs=vector<string>;
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rep(i,n) range(i,0,n)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define range(i,a,n) for(int i=(a);i<(n);i++)

#define LINF    	((ll)1ll<<60)
#define INF     	((int)1<<30)
#define EPS     	(1e-9)
#define MOD     	(1000000007ll)
#define fcout(a)	cout<<setprecision(a)<<fixed
#define fs 			first
#define sc			second
#define PI			3.1415926535897932384

int dx[]={1,0,-1,0,1,-1,-1,1},dy[]={0,1,0,-1,1,1,-1,-1};
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool a=1; for(auto s:t){os<<(a?"":" ")<<s; a=0;} return os;}
void OUT(bool b){cout<<(b?"YES":"NO")<<endl;}
void Out(bool b){cout<<(b?"Yes":"No")<<endl;}
void out(bool b){cout<<(b?"yes":"no")<<endl;}
int popcnt(int x) { return __builtin_popcount(x); }
int popcnt(ll x) { return __builtin_popcountll(x); }
int clz(int x) { return 31 - __builtin_clz(x); }
int clz(ll x) { return 63 - __builtin_clzll(x); }
int ctz(int x) { return __builtin_ctz(x); }
int ctz(ll x) { return __builtin_ctzll(x); }

class compare {
public:
	bool operator()(tuple<ll, ll> a, tuple<ll, ll> b) {
		return (get<1>(a) > get<1>(b));
	}
};

void solve(){
	string a,b;
	cin>>a>>b;
	vvi dp(a.size()+1,vi(b.size()+1));
	rep(i,a.size()){
		rep(j,b.size()){
			if(a[i]==b[j])dp[i+1][j+1]=dp[i][j]+1;
			else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
		}
	}
	cout<<dp[a.size()][b.size()]<<endl;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	rep(i,n)solve();
}