The main idea of the topic ： Given n The number and two lengths are n*5 Sequence , Every number happens to be 5 Time , Find the... Of two sequences LCS

n<=20000. The length of the sequence is 10W. Simple O(n^2) It's bound to time out

So we think about LCS Some properties of

LCS The decision +1 Is the condition of a[i]==b[j] So we recorded a Of every number in the sequence 5 A place

Sweep it b[i] For each of these b[i] find b[i] stay a Medium 5 A place this 5 Every one of the positions f[pos] Values can be b[i] to update So I found f[1] To f[pos-1] The maximum of +1 to update f[pos] Can

This is maintained with a tree array Time complexity O(nlogn)

A very difficult question It's just not hard to write

```#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 200200
using namespace std;
int n,ans,a[M*5],b[M*5],c[M*5],f[M*5],pos[M][6];
void Update(int x,int y)
{
for(;x<=n*5;x+=x&-x)
c[x]=max(c[x],y);
}
int Get_Ans(int x)
{
int re=0;
for(;x;x-=x&-x)
re=max(re,c[x]);
return re;
}
int main()
{
int i,j;
cin>>n;
for(i=1;i<=n*5;i++)
{
scanf("%d",&a[i]);
pos[ a[i] ][ ++pos[a[i]][0] ]=i;
}
for(i=1;i<=n*5;i++)
scanf("%d",&b[i]);
for(i=1;i<=n*5;i++)
{
for(j=5;j;j--)
{
int k=pos[b[i]][j];
f[k]=max( f[k] , Get_Ans(k-1)+1 );
Update(k,f[k]);
ans=max(ans,f[k]);
}
}
cout<<ans<<endl;
}
```

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