Two Rabbits

Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon, they planned to play a game with some stones. There were n stones on the ground and they were arranged as a clockwise ring. That is to say, the first stone was adjacent to the second stone and the n-th stone, and the second stone is adjacent to the first stone and the third stone, and so on. The weight of the i-th stone is ai.

The rabbits jumped from one stone to another. Tom always jumped clockwise, and Jerry always jumped anticlockwise.

At the beginning, the rabbits both choose a stone and stand on it. Then at each turn, Tom should choose a stone which have not been stepped by itself and then jumped to it, and Jerry should do the same thing as Tom, but the jumping direction is anti-clockwise.

For some unknown reason, at any time , the weight of the two stones on which the two rabbits stood should be equal. Besides, any rabbit couldn't jump over a stone which have been stepped by itself. In other words, if the Tom had stood on the second stone, it cannot jump from the first stone to the third stone or from the n-the stone to the 4-th stone.

Please note that during the whole process, it was OK for the two rabbits to stand on a same stone at the same time.

Now they want to find out the maximum turns they can play if they follow the optimal strategy.

Input

The input contains at most 20 test cases.
For each test cases, the first line contains a integer n denoting the number of stones.
The next line contains n integers separated by space, and the i-th integer ai denotes the weight of the i-th stone.(1 <= n <= 1000, 1 <= ai <= 1000)
The input ends with n = 0.

Output

For each test case, print a integer denoting the maximum turns.

Sample

Inputcopy	Outputcopy
1 1 4 1 1 2 1 6 2 1 1 2 1 3 0	1 4 5

Hint

For the second case, the path of the Tom is 1, 2, 3, 4, and the path of Jerry is 1, 4, 3, 2.
For the third case, the path of Tom is 1,2,3,4,5 and the path of Jerry is 4,3,2,1,5.

题目大意

就是有二个兔子，一个顺时针走，一个逆时针走，他们走的路线是要相同的。

比如例子中的2 1 1 2 1 3

 Tom走的是2 1 1 2 1（1,2,3,4,5）；Jerry走的是2 1 1 2 1（4,3,2,1,5）

还有就是不能跨过你已经踩过了的，比如你在1,2踩过了，你就不能跳了

现在需要你找出一个最长的路线。

相当于是找一条最长的回文子串。

解析

这一题我用的是DP

DP[i][j]:表示从i到j的最长的回文子串。

最优解公式：if(a[i]==a[j]) dp[i][j]=dp[i+1][j-1]+2;

                     else dp[i][j]=max(dp[i+1][j],dp[i][j-1]);

当（i,j）这个区间里的首部a[i]==尾部a[j]时，(i,j)里面的最长回文子串等于（i+1,j-1）里面最长的回文子串+2；

当不相等的时候，取（i+1,j）和(i,j-1)里面回文子串的最大值。

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<vector>
#include<map>
#include<cstring>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<fstream>
using namespace std;
typedef long long ll;
int dx[8]={-2,-1,1,2,2,1,-1,-2};
int dy[8]={1,2,2,1,-1,-2,-2,-1};
int dx1[4]={-1,0,1,0};
int dy1[4]={0,1,0,-1};
const int mod= 1000000007;
int gcd(int a,int b){
  return(b?gcd(b,a%b):a);
}
int lcm(int a,int b){
  return(a*b/gcd(a,b));
}
int dp[1005][1005];
int main(){
    int n;
    while(cin>>n,n){
        memset(dp,0,sizeof(dp));
        vector<int> a(n+5);
        for(int i=1;i<=n;i++){
            cin>>a[i];
            dp[i][i]=1;
        }
        for(int l=1;l<n;l++)
        for(int i=1;i<=n;i++){
            int j=i+l;
            if(j>n){
                break;
            }
            if(a[i]==a[j]){
                dp[i][j]=dp[i+1][j-1]+2;
            }
            else{
                dp[i][j]=max(dp[i+1][j],dp[i][j-1]);
            }
        }
        int m=0;
        for(int i=1;i<=n;i++){
            m=max(m,dp[1][i]+dp[i+1][n]);
        }
        cout<<m<<endl;
    }
}