题意

找如下子串的个数：
(l,r)是回文串，并且(l,(l+r)/2)也是回文串

思路

本来写了个回文树+dfs+hash，由于用了map所以T了
后来发现既然该子串和该子串的前半部分都是回文串，所以该子串的前半部分和后半部分是本质相同的！
于是这个log就去掉了

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 8e5+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

const ull base = 201326611;

ull po[maxn],ha[maxn];

ull getHash(int l, int r){
    return ha[r]-ha[l-1]*po[r-l+1];
}
int ans[maxn];
int id[maxn];
struct pam{
    int nx[maxn][26],num[maxn],cnt[maxn],fail[maxn];
    int len[maxn],s[maxn],p,lst,n;
    int newNode(int x){
        mem(nx[p],0);
        cnt[p]=num[p]=0;
        len[p]=x;
        return p++;
    }
    void init(){
        p=0;newNode(0);newNode(-1);
        lst=0;n=0;s[0]=-1;fail[0]=1;
    }
    int getFail(int x){
        while(s[n-len[x]-1]!=s[n])x=fail[x];
        return x;
    }
    void add(int x){
        x-='a';
        s[++n]=x;
        int cur=getFail(lst);
        if(!(lst=nx[cur][x])){//产生新节点
            int now = newNode(len[cur]+2);
            fail[now]=nx[getFail(fail[cur])][x];
            nx[cur][x]=now;
            num[now]=num[fail[now]]+1;
            lst=now;

            id[now]=n;
        }
        cnt[lst]++;
    }
    void count(){
        for(int i = p-1; i >= 0; i--)cnt[fail[i]]+=cnt[i];
        for(int i = 2; i < p; i++){
            int l = id[i]-len[i]+1;
            int r = id[i];
            int mid = (l+r)>>1;
            if(len[i]%2==1&&getHash(l,mid)==getHash(mid,r))ans[len[i]]+=cnt[i];
            if(len[i]%2==0&&getHash(l,mid)==getHash(mid+1,r))ans[len[i]]+=cnt[i];
        }
    }
}pam;
int n;
char s[maxn];
int main(){
    po[0]=1;
    for(int i = 1; i <= 3e5+10; i++){
        po[i]=po[i-1]*base;
    }
    while(~scanf("%s",s+1)){
        n=strlen(s+1);
        pam.init();
        for(int i = 1; i <= n; i++){
            ans[i]=0;
            ha[i]=ha[i-1]*base+s[i]-'a'+1;
            pam.add(s[i]);
        }pam.count();
        for(int i = 1; i <= n; i++){
            printf("%d", ans[i]);
            if(i!=n)printf(" ");
        }printf("\n");
    }
    return 0;
}
/*

 */