# Harry and magic string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 223    Accepted Submission(s): 110

Problem Description

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 (where a1 is the beginning index of  is the ending index of x.  as the same of y), if both x and y are palindromes and  then we consider (x, y) to be a disjoint palindrome substring pair of T.

Input

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].

Output

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

Sample Input

```aca
aaaa```

Sample Output

```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;
}
{
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--)
{
}
tree.init();
LL ans=0;
for(int i=0;i<len;i++)
{
}
printf("%lld\n",ans);
}
return 0;
}```

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