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

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

 


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;
    }
    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;
}