It is easy to find that the length of the palindrome that meets the requirements must be a multiple of 4, and there is a suffix palindrome with half the length.
Memorize the search after building the fail tree
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;;
const int SZ = 26;///字符集
const int maxn = 5e5 + 6;
struct PAM {
struct PamNode{
int fail,trans[SZ],sz,len,num;}pam[maxn];
int tot;char s[maxn];
vector<int>G[maxn];
void init() {
tot=1;
pam[0].fail=1;pam[0].len=0;
pam[1].fail=1;pam[1].len=-1;
memset(pam[0].trans,0,SZ*sizeof (int));
memset(pam[1].trans,0,SZ*sizeof (int));
}
inline int newnode(int len) {
tot++;
memset(pam[tot].trans,0,SZ*sizeof (int));
pam[tot].len=len;pam[tot].fail=0;
pam[tot].sz=0;pam[tot].num=0;
return tot;
}
inline int getfail(int i,int u) {
while(s[i-pam[u].len-1]^s[i]) u=pam[u].fail;
return u;
}
inline int append(int i,int u) {
int c=s[i]-'a';
int fa=getfail(i,u);
u=pam[fa].trans[c];
if(!u) {
int z=newnode(pam[fa].len+2);
int w=getfail(i,pam[fa].fail);
pam[z].fail=pam[w].trans[c];
G[pam[z].fail].pb(z);
u=pam[fa].trans[c]=z;///注意这里要后更新，否则上面getfail时可能导致死循环
pam[z].num=pam[pam[z].fail].num+1;
}
pam[u].sz++;return u;
}
void calu() {
for(int i=tot,fail;i>1;i--) {
fail=pam[i].fail;
pam[fail].sz+=pam[i].sz;
}
}
int vis[maxn],ans;
void dfs(int u) {
//cout<<u<<','<<pam[u].len<<endl;
if(u>1) vis[pam[u].len]=1;
if(u>1&&pam[u].len%4==0) {
if(vis[pam[u].len/2]) ans=max(ans,pam[u].len);
}
for(auto i:G[u]) dfs(i);
vis[pam[u].len]=0;
}
void sol() {
ans=0;
dfs(0);dfs(1);
printf("%d\n",ans);
}
}pa;
int n,_,cs=0;
int main() {
scanf("%d",&n);pa.init();
scanf("%s",pa.s+1);
for(int i=1,last=0;i<=n;i++) {
last=pa.append(i,last);
}
pa.sol();
}