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|1using 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;
}
/**/