Harry got a string T, he wanted to know the number of T’s disjoint palindrome substring pairs. A string is considered to be palindrome if and only if it reads the same backward or forward. For two substrings of T:x=T[a1…b1],y=T[a2…b2](where a1 is the beginning index of x,b1 is the ending index of x. a2,b2 as the same of y), if both x and y are palindromes and b1<a2 or b2<a1 then we consider (x, y) to be a disjoint palindrome substring pair of T.

There are several cases.
For each test case, there is a string T in the first line, which is composed by lowercase characters. The length of T is in the range of [1,100000].

For each test case, output one number in a line, indecates the answer.

aca
aaaa

3
15
求一个字符串中所有不相交的回文串对
回文树，先倒着扫一遍，再正着扫一遍
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
const int MAX=1e5+5;
char str[MAX];
struct Tree
{
	int next[MAX][26];
	int fail[MAX];
	LL num[MAX];
	int len[MAX];
	int s[MAX];
	int last;
	int n;
	int p;
	int new_node(int x)
	{
		memset(next[p],0,sizeof(next[p]));
		num[p]=0;
		len[p]=x;
		return p++;
	}
	void init()
	{
		p=0;
		new_node(0);
		new_node(-1);
		last=0;
		n=0;
		s[0]=-1;
		fail[0]=1;
	}
	int get_fail(int x)
	{
		while(s[n-len[x]-1]!=s[n])
			x=fail[x];
		return x;
	}
	LL add(int x)
	{
		x-='a';
		s[++n]=x;
		int cur=get_fail(last);
		if(!(last=next[cur][x]))
		{
			int now=new_node(len[cur]+2);
			fail[now]=next[get_fail(fail[cur])][x];
			next[cur][x]=now;
			num[now]=num[fail[now]]+1;
			last=now;
		}
		return num[last];
	}
}tree;
LL sum[MAX];
int main()
{
    while(scanf("%s",str)!=EOF)
	{
		int len=strlen(str);
		sum[len]=0;
		tree.init();
		for(int i=len-1;i>=0;i--)
		{
           sum[i]=sum[i+1]+tree.add(str[i]);
		}
		tree.init();
		LL ans=0;
		for(int i=0;i<len;i++)
		{
            ans+=(LL)tree.add(str[i])*sum[i+1];
		}
		printf("%lld\n",ans);
	}
	return 0;
}