动态规划求回文长度
设len是该环上最长的回文串的长度,当len等于n时,ans=len;当len<n时,ans=len+1;
#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
int n;
int a[2100];
int dp[2100][2100];
int main()
{
while(cin>>n&&n)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
cin>>a[i];
a[i+n]=a[i];
dp[i][i]=dp[i+n][i+n]=1;
}
for(int len=2;len<=n;len++)
for(int i=1;i+len-1<2*n;i++)
{
int k=i+len-1;
if(a[i]==a[k])
dp[i][k]=dp[i+1][k-1]+2;
else
dp[i][k]=max(dp[i][k],max(dp[i+1][k],dp[i][k-1]));
}
int ans=0;
for(int i=1;i<=n;i++)
ans=max(ans,max(dp[i][i+n-1],dp[i][i+n-2]+1));
cout<<ans<<endl;
}
return 0;
}