题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5371

题意：求形如ABA（B与A对称0）的串的最大长度。

解法：先用manacher预处理出以每个点为中心的最长回文子串长度，枚举第二个部分即B的起点，再枚举串的长度，发现可行则更新答案（代码中非常清楚了）。可以用已经得到的ans剪枝。我求答案的时候用i++,j--枚举T了，改成i+=2,j-=2过了。。


#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5+10;
typedef long long LL;
struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) return -1;
        return buf[pos ++];
    }
    inline int xuint()
    {
        int c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) * s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(LL x)
    {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;
int s[maxn], str[maxn];
int len1,len2,p[maxn],ans;
void init(){
    str[0]=-1;
    str[1]=-1;
    for(int i=0; i<len1; i++){
        str[i*2+2]=s[i];
        str[i*2+3]=-1;
    }
    len2=len1*2+2;
    str[len2]=-1;
}
void manacher(){
    int id=0,mx=0;
    for(int i=1; i<len2; i++){
        if(mx>i) p[i]=min(p[2*id-i],mx-i);
        else p[i]=1;
        for(;str[i+p[i]]==str[i-p[i]];p[i]++);
        if(p[i]+i>mx){
            mx=p[i]+i;
            id=i;
        }
    }
}
int main()
{
    int T,n,ks=0;
    T = io.xint();
    while(T--){
        n = io.xint();
        for(int i=0; i<n; i++){
            s[i] = io.xint();
        }
        len1=n;
        init();
        manacher();
        ans=0;
        for(int i=1; i<len2; i+=2){
            for(int j=p[i]+i-1; j-i>ans; j-=2){
                if(j-i+1<=p[j]){
                    ans = max(ans, j-i);
                    break;
                }
            }
        }
        printf("Case #%d: %d\n", ++ks, ans/2*3);
    }
    return 0;
}