Given two strings A and B, find the longest common subsequence of A and B (the subsequence does not need to be continuous).

For example, two strings are:

abcicba

abdkscab

ab is a subsequence of two strings, so is abc and abca. ABCA is the longest subsequence of the two strings.

Retract

# input

```Line 1: String A
Line 2: String B
(Length of A,B <= 1000)```

# output

`Output the longest subsequence, if there are more than one, output one at will.`

# sample input

```abcicba
abdkscab```

# sample output

`abca`

The idea is to find the longest common subsequence first, and then to find the matching point according to the path inverse.

```#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//

#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0)
#define file  freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//

#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
//--------------------------------constant----------------------------------//

#define INF  0x3f3f3f3f
#define maxn  1005
#define esp  1e-9
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
//------------------------------Dividing Line--------------------------------//
string a,b,ans;
int dp[maxn][maxn];
int main()
{
ans.clear();
cin>>a>>b;
a=' '+a;
b=' '+b;
int la=a.size();
int lb=b.size();
for(int i=1;i<la;i++)
{
for(int j=1;j<lb;j++)
{
if(a[i]==b[j])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
int m=la-1,n=lb-1;
while(dp[m][n])
{
if(dp[m][n]==dp[m-1][n]) m--;
else if(dp[m][n-1]==dp[m][n]) n--;
else
{
if(a[m]!=' ')
ans=a[m]+ans;
m--,n--;
}
}
cout<<ans<<endl;
}
```