顯然答案應該是由單串以某位置為中心的極長回文串繼續在另一個串裏拓展得到的。枚舉中間位置二分答案，哈希判斷即可。註意考慮清楚怎麽處理偶回文，比如像manacher一樣加分隔符。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
#define ul unsigned long long
char
 
 getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while
 
 (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,ans;
ul ha[N],rha[N],hb[N],rhb[N],p[N];
char a[N],b[N],s1[N],s2[N];
ul hash(ul *a,int l,int r,int op)
{
    return a[r]-a[l-op]*p[abs(r-l)+1];
}
void solve(ul *a,ul *ra,ul *b,ul *rb,int op,int n)
{
    
 
for (int i=1;i<=n;i++)
    {
        int l=0,r=min(n-i,i-1),len=0;
        while (l<=r)
        {
            int mid=l+r>>1;
            if (hash(a,i-mid,i,1)==hash(ra,i+mid,i,-1)) len=mid,l=mid+1;
            else r=mid-1;
        }
        ans=max(ans,len);
        l=1,r=min(i-len+1,n-i-len);
        while (l<=r)
        {
            int mid=l+r>>1;
            if (op?hash(b,i-len-mid+1,i-len,1)==hash(ra,i+len+mid,i+len+1,-1):hash(a,i-len-mid,i-len-1,1)==hash(rb,i+len+mid-1,i+len,-1)) ans=max(ans,len+mid),l=mid+1;
            else r=mid-1;
        }
    }
}
void work(int n)
{
    n=n*2+2;
    for (int i=1;i<=n;i++) ha[i]=ha[i-1]*509+a[i];
    for (int i=n;i>=1;i--) rha[i]=rha[i+1]*509+a[i];
    for (int i=1;i<=n;i++) hb[i]=hb[i-1]*509+b[i];
    for (int i=n;i>=1;i--) rhb[i]=rhb[i+1]*509+b[i];
    p[0]=1;for (int i=1;i<=n;i++) p[i]=p[i-1]*509;
    solve(ha,rha,hb,rhb,0,n),solve(hb,rhb,ha,rha,1,n);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4755.in","r",stdin);
    freopen("bzoj4755.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    scanf("%s",s1+1),scanf("%s",s2+1);
    for (int i=n;i>=1;i--) a[i*2+2]=‘$‘,a[i*2+1]=s1[i];a[2]=‘$‘;a[1]=‘#‘;
    for (int i=n;i>=1;i--) b[i*2+1]=‘$‘,b[i*2]=s2[i];b[1]=‘$‘;b[n*2+2]=‘@‘;
    work(n);
    for (int i=n;i>=1;i--) a[i*2+1]=‘$‘,a[i*2]=s1[i];a[1]=‘$‘;a[n*2+2]=‘@‘;
    for (int i=n;i>=1;i--) b[i*2+2]=‘$‘,b[i*2+1]=s2[i];b[2]=‘$‘;b[1]=‘#‘;
    work(n);
    cout<<ans;
    return 0;
}

BZOJ4755 JSOI2016扭動的回文串（二分答案+哈希）