题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6599

利用回文树找出所有的回文串(记录出现的位置)
然后利用哈希判断回文串的一半是否也是回文串即可

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int maxn = 300010;
const ull base = 131;

int n, tot;
int tail, last;

ull has[maxn], rhash[maxn], po[maxn];
int ans[maxn];

struct PAM{
	int len, fail, son[30], sz, pos;
}t[maxn];

char s[maxn];

int Get_Fail(int x){
	while(s[tail - t[x].len - 1] != s[tail]) x = t[x].fail;
	return x;
}

void Extend(int x){
	int cur = Get_Fail(last);
	if(!t[cur].son[x]){
		++tot;
		t[tot].len = t[cur].len + 2;
		t[tot].fail = t[Get_Fail(t[cur].fail)].son[x];
		t[tot].pos = tail;
		t[cur].son[x] = tot;
	}
	++t[t[cur].son[x]].sz;
	last = t[cur].son[x];
}

void count(){
	for(int i = tot ; i >= 2 ; --i){
		t[t[i].fail].sz += t[i].sz;
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	po[0] = 1;
	for(int i = 1 ; i <= 300000 ; ++i) po[i] = po[i - 1] * base;
	while(scanf("%s", s + 1) != EOF){
		memset(t, 0, sizeof(t));
		memset(ans, 0, sizeof(ans));
		memset(rhash, 0, sizeof(rhash));
		memset(has, 0, sizeof(has));
		n = strlen(s + 1);
		
		for(int i = 1 ; i <= n ; ++i) has[i] = has[i - 1] * base + s[i] - 'a';
		rhash[n + 1] = 0;
		for(int i = n ; i >= 1 ; --i) rhash[i] = rhash[i + 1] * base + s[i] - 'a';
		
		t[0].len = 0, t[1].len = -1;
		t[0].fail = t[1].fail = 1;
		last = 1; tot = 1;
		
		for(tail = 1 ; tail <= n ; ++tail){
			Extend(s[tail] - 'a');
		}
		
		count();
		
		for(int i = 2 ; i <= tot ; ++i){
			int len2 = (t[i].len & 1) ? t[i].len / 2 + 1 : t[i].len / 2;
			int len = len2 / 2;
			int mid = t[i].pos - ((t[i].len & 1) ? ((t[i].len / 2) + 1) : t[i].len / 2) + 1;
//			printf("%d %d %d\n", t[i].pos, t[i].len, mid);
			
			int HR = has[t[i].pos] - has[t[i].pos - len] * po[len];
			int HL = rhash[mid] - rhash[mid + len] * po[len];
			if(HL == HR){
				ans[t[i].len] += t[i].sz;
			} 
		}
		
		for(int i = 1 ; i <= n ; ++i) {
			if(i == 1) printf("%d", ans[i]);
			else printf(" %d", ans[i]);
		}
		puts("");
	}
	return 0;
}