这道题可以$O(nlogn)$，当然也可以$O(n)$做啦$qwq$

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$O(nlogn)$的思路是枚举每个回文中心，通过哈希预处理出前缀和后缀哈希值备用，然后二分回文串的长度，具体的就是判断在长度范围内，前缀哈希值和后缀哈希值是否相等。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ll unsigned long long
#define R register int
using namespace std;
namespace Fread {
    static char B[1<<15],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
        do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
    }
}using Fread::g;
const int B=131,N=1000010;
int t,ans;
ll h1[N],h2[N],p[N];
char s[N];
inline ll H1(int l,int r) {return h1[r]-h1[l-1]*p[r-l+1];}
inline ll H2(int l,int r) {return h2[l]-h2[r+1]*p[r-l+1];}
signed main() {
#ifdef JACK
    freopen("NOIPAK++.in","r",stdin);
#endif
    p[0]=1; for(R i=1;i<=N-10;++i) p[i]=p[i-1]*B;
    while(1) { ans=0; ++t; 
        scanf("%s",s+1); R len=strlen(s+1);
        if(len==3&&s[1]=='E'&&s[2]=='N'&&s[3]=='D') break;
        for(R i=1;i<=len;++i) h1[i]=h1[i-1]*B+(s[i]^48)+1;
        for(R i=len;i>=1;--i) h2[i]=h2[i+1]*B+(s[i]^48)+1;
        for(R p=1;p<=len;++p) {
            register int l=1,r=min(p-1,len-p);
            while(l<r) {
                register int md=l+r+1>>1;
                if(H1(p-md,p-1)==H2(p+1,p+md)) l=md;
                else r=md-1;
            } ans=max(ans,l*2+1);
            l=1,r=min(p-1,len-p+1);
            while(l<r) {
                register int md=l+r+1>>1;
                if(H1(p-md,p-1)==H2(p,p+md-1)) l=md;
                else r=md-1;
            } ans=max(ans,l*2);
        } printf("Case %d: %d\n",t,ans);
    }
}

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还有一个$Manacher$算法，可以在$O(n)$时间里解决这个问题，详见这位大佬的博客

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ll long long
#define R register int
using namespace std;
namespace Fread {
    static char B[1<<15],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
        do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
    }
}using Fread::g;
const int N=1000010;
char s[N<<1],str[N<<1];
int len[N<<1],l;
inline void getstr() {
    R k=0; str[k]='$';
    for(R i=0;i<=l;++i) str[++k]='#',str[++k]=s[i];
    str[++k]='#'; l=k;
}
inline void Manacher() {
    getstr(); R mx=0,id;
    for(R i=1;i<l;++i) {
        if(mx>i) len[i]=min(len[2*id-i],mx-i);//先找对称的位置，和右边界取一个min 
        else len[i]=1;//不包含直接设为1 
        while(str[i+len[i]]==str[i-len[i]]) ++len[i];//暴力匹配 
        if(len[i]+i>mx) mx=len[i]+i,id=i;//更新mx和id 
    }
}
signed main() {
#ifdef JACK
    freopen("NOIPAK++.in","r",stdin);
#endif
    R t=0; while(1) { memset(len,0,sizeof(len));  ++t;
        scanf("%s",s); l=strlen(s);
        if(l==3&&s[0]=='E'&&s[1]=='N'&&s[2]=='D') break;
        Manacher(); R ans=1; for(R i=1;i<l;++i) ans=max(ans,len[i]); 
        printf("Case %d: %d\n",t,ans-1);
    } 
}

洛谷上也有$manacher$的模板，可以水一下；

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2019.06.10