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)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)S[i]=S[2n−i]=S[2n+i−2](1≤i≤n).For example, abcbabcabcbabc is one-and-half palindromic string, and abccbaabcabccbaabc 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 500000500000), 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
Hint
In the example input, there are two substrings which are one-and-half palindromic strings, $abab$ and $abcbabc$.
题意：给你t个字符串，定义1.5倍回文串为S[i]=S[2n−i]=S[2n+i−2](1≤i≤n)S[i]=S[2n−i]=S[2n+i−2](1≤i≤n).说简单点就是两个点，i
和j都都为回文串的中点其中每个串的长度为(j-i)*2，比如abab，abcbabc，问你这个5e5的串里面有多少个这样的回文串。
思路：这什么1.5倍回文串，暴力想都别想，但是我们可以用Manacher算法预处理一下，把每个字符组成回文串想两边可以延伸的长度len[]处理出来，这样直接找一对i，j使i+len[i]>=j&&j-len[i]<=i,这样是就是一个1.5倍回文串了，但是如果暴力枚举一个i的len的话那么复杂度又被拉回到了n*n了，一定不能过，开始的时候除了怎么找一个i符合要求的j其他的思路都想到了，还是自己太菜了，线段树，优先队列，贪心，都想过，实在想不出来了，最后去看了下题解，居然是树状数组，知道要用树状数组后自己也想了好长一段时间，就是想不到，怎么去解决这个问题，最后没办法了，仔细的看看别人的代码，才明白了树状数组是怎么处理的。先枚举记录一个字符可以延伸到前面的那个地方cnt-maxr，在记录那个点时谁能及的，然后记录开始遍历每一个字符，把所有到该点的点都加入到树状数组，然后计算sum(i+maxr)-sum(i)，这样就保证了i+len[i]>=j&&j-len[i]<=i十分的巧妙，想了好久的确没想到。
代码：
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<set>
#include<map>
#define ll long long
#define qq printf("QAQ\n");
using namespace std;
const int maxn=1e6+5;
const int inf=0x3f3f3f3f;
const ll linf=8e18+9e17;
const int mod=1e9+7;
const double e=exp(1.0);
const double pi=acos(-1);
const double eps=1e-6;
char s[maxn],ss[maxn];
int len[maxn],sum[maxn],maxs[maxn];
vector<int>v[maxn];
int lowbit(int x)
{
return x & - x;
}
void updata(int x,int data)
{
while(x<maxn)
{
sum[x]+=data;
x+=lowbit(x);
}
}
int getsum(int x)
{
int ans=0;
while(x>0)
{
ans+=sum[x];
x-=lowbit(x);
}
return ans;
}
int Manacher()
{
int ls=strlen(s),lss=1;
ss[0]='!';
for(int i=0;i<ls;i++)
{
ss[lss++]='#';
ss[lss++]=s[i];
}
ss[lss++]='#';
ss[lss++]='@';
int maxr=-1,j=-1,mid;
for(int i=1;i<lss;i++)
{
if(maxr>i)len[i]=min(len[mid*2-i],maxr-i);
else len[i]=1;
while(ss[i-len[i]]==ss[i+len[i]])len[i]++;
if(len[i]+i>maxr)maxr=len[i]+i,mid=i;
}
return lss;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s",s);
memset(sum,0,sizeof sum);
int l=Manacher(),cnt=1;
ll ans=0;
for(int i=2;i<l;i+=2,cnt++)
{
maxs[cnt]=len[i]/2-1;
v[cnt-maxs[cnt]].push_back(cnt);
}
for(int i=1;i<=cnt;i++)
{
for(int j=0;j<v[i].size();j++)
updata(v[i][j],1);
ans+=(ll)(getsum(i+maxs[i])-getsum(i));
v[i].clear();
}
printf("%lld\n",ans);
}
return 0;
}