This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

InputEach sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.Outputoutput print L - the length of the greatest common increasing subsequence of both sequences.Sample Input

1

5
1 4 2 5 -12
4
-12 1 2 4

Sample Output

2

注意一下控制格式

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<iomanip>
 6 #include<string.h>
 7 using namespace std;
 8 
 9 int a[550];
10 int b[550];
11 int dp[550];
12 
13 int main()
14 {
15     std::ios::sync_with_stdio(false);
16     cin.tie(0);
17     cout.tie(0);
18     int tt;
19     cin>>tt;
20     while(tt--)
21     {
22         memset(a,0,sizeof(a));
23         memset(b,0,sizeof(b));
24         memset(dp,0,sizeof(dp));
25         int p,q;
26         cin>>p;
27         for(int i=0;i<p;i++)
28             cin>>a[i];
29         cin>>q;
30         for(int i=0;i<q;i++)
31             cin>>b[i];
32         int maxx;
33         for(int i=0;i<p;i++)
34         {
35             maxx=0;//每次都要恢复
36             for(int j=0;j<q;j++)
37             {
38                 if(a[i]>b[j])
39                     maxx=max(dp[j],maxx);
40                 else if(a[i]==b[j])
41                     dp[j]=maxx+1;
42             }
43         }
44         int ans=0;
45         for(int i=0;i<q;i++)
46             ans=max(ans,dp[i]);
47         if(tt)
48             cout<<ans<<endl<<endl;
49         else
50             cout<<ans<<endl;
51     }
52     return 0;
53 }