题意：给定一个圆形的环，有两个只兔子，一只顺时针跳，一个逆时针，但每次跳到的石头必须一样，问你最多能跳多少轮。
析：本来以为是LCS呢，把那个序列看成一个回文，然后就能做了，但是时间受不了。其实是一个区间DP，dp[i[j] 表示从 i 到 j 中最长的回文数。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int dp[maxn][maxn];
int main(){
while(scanf("%d", &n) == 1 && n){
for(int i = 0; i < n; ++i) scanf("%d", a+i);
memset(dp, 0, sizeof dp);
for(int i = 0; i < n; ++i) dp[i][i] = 1;
for(int i = n-2; i >= 0; --i)
for(int j = i+1; j < n; ++j)
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 ans = 0;
for(int i = 0; i < n; ++i) ans = Max(ans, dp[0][i]+dp[i+1][n-1]);
printf("%d
", ans);
}
return 0;
}