簡單，數據為 1e3

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
char a[N], b[N];
int n, m, f[N][N];
void solve(){
	cin >> n >> m >> a + 1 >> b + 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++){
			f[i][j] = max(f[i - 1][j], f[i][j - 1]);
			if (a[i] == b[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
		}
	cout << f[n][m] << "\n";
}
int main(){
	solve();
	return 0;
}

acwing模板：https://www.acwing.com/problem/content/899/

數據為 1e5

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 1e5 + 10;  //數組大小
int n, a[maxn], b[maxn], f[maxn], p[maxn];
int main(){
	cin >> n;
	for (int i = 1; i <= n; i++){
		scanf("%d", &a[i]);
		p[a[i]] = i;  //將第二個序列中的元素映射到第一個中
	}
	for (int i = 1; i <= n; i++){
		scanf("%d", &b[i]);
		f[i] = INF;
	}
	int len = 0;
	f[0] = 0;
	for (int i = 1; i <= n; i++){
		if (p[b[i]] > f[len]) f[++len] = p[b[i]];
		else {
			int l = 0, r = len;
			while (l < r){
				int mid = (l + r) >> 1;
				if (f[mid] > p[b[i]]) r = mid;
				else l = mid + 1;
			}
			f[l] = min(f[l], p[b[i]]);
		}
	}
	cout << len << "\n";
	return 0;
}

洛谷模板：https://www.luogu.com.cn/problem/P1439