Palindrome

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 426    Accepted Submission(s): 163

Problem Description

Alice like strings, especially long strings. For each string, she has a special evaluation system to judge how elegant the string is. She defines that a string

S[1..3n−2](n≥2) is one-and-half palindromic if and only if it satisfies S[i]=S[2n−i]=S[2n+i−2](1≤i≤n).For example, abcbabc is one-and-half palindromic string, and abccbaabc is not. Now, Alice has generated some long strings. She ask for your help to find how many substrings which is one-and-half palindromic.

 

 

Input

The first line is the number of test cases. For each test case, there is only one line containing a string(the length of strings is less than or equal to

500000), this string only consists of lowercase letters.

 

 

Output

For each test case, output a integer donating the number of one-and-half palindromic substrings.

 

 

Sample Input

1 ababcbabccbaabc

 

 

Sample Output

2

/*************************************************************************    > File Name: A.cpp    > Author: LyuCheng    > Created Time: 2017-12-01 21:54    > Description:         题意：给你一个字符串问你有多少个子串满足s[i]=s[2*n-i]=s[2*n+i-2]        思路：满足条件的子串实际上是关于两点回文的一个子串,并且这个回文串               一定是奇数回文,用马拉车算法求出以每个点的回文半径，如果满足题            目的条件，设i
      
       <=pos[i]&&i-j<=pos[j]            用树状数组记录一下，然后统计 ************************************************************************/#include 
       
        #define MAXN 567890#define lowbit(x) x&(-x)#define LL long longusing namespace std;int t;int n;char str[MAXN<<1];char tmp[MAXN<<1];int Len[MAXN<<1];int pos[MAXN<<1];int c[MAXN<<1];LL res;vector
        
         v[MAXN<<1];inline void update(int x){    while(x
         
          0){        s+=c[x];        x-=lowbit(x);    }    return s;}int cal(char *st){    int i,len=strlen(st);    tmp[0]='@';//字符串开头增加一个特殊字符，防止越界    for(i=1;i<=2*len;i+=2){        tmp[i]='#';        tmp[i+1]=st[i/2];    }    tmp[2*len+1]='#';    tmp[2*len+2]='$';//字符串结尾加一个字符，防止越界    tmp[2*len+3]=0;    return 2*len+1;//返回转换字符串的长度}//Manacher算法计算过程int manacher(char *st,int len){     int mx=0,ans=0,po=0;//mx即为当前计算回文串最右边字符的最大值     for(int i=1;i<=len;i++){         if(mx>i)             Len[i]=min(mx-i,Len[2*po-i]);//在Len[j]和mx-i中取个小         else             Len[i]=1;//如果i>=mx，要从头开始匹配         while(st[i-Len[i]]==st[i+Len[i]])             Len[i]++;         if(Len[i]+i>mx){
    //若新计算的回文串右端点位置大于mx，要更新po和mx的值             mx=Len[i]+i;             po=i;         }         ans=max(ans,Len[i]);     }     return ans-1;//返回Len[i]中的最大值-1即为原串的最长回文子串额长度}inline void init(){    memset(str,'\0',sizeof str);    memset(tmp,'\0',sizeof tmp);    memset(c,0,sizeof c);    memset(pos,0,sizeof pos);    for(int i=0;i