D. Palindrome Degree

题目连接：

http://www.codeforces.com/contest/7/problem/D

Description

String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.

Let’s call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, “abaaba” has degree equals to 3.

You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.

Input

The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.

Output

Output the only number — the sum of the polindrome degrees of all the string’s prefixes.

Sample Input

a2A

Sample Output

1

Hint

题意

如果一个数是k回文串的话，那么他是回文串，且他的前半缀是k-1回文串，他的后半缀也是k-1回文串。

然后问你这个所有前缀的回文串等级的和是多少

题解：

跑的时候，维护这个前缀正着的的hash值，这个前缀倒着的hash值。

如果这两个hash值相同的话，说明这个串是一个回文串，那么他的等级d[i]=d[i/2]+1

然后跑一遍就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e7+5;
char s[maxn];
int dp[maxn];
long long p1=131,t1=1;
long long p2=127,t2=1;
long long h1=0,rh1=0;
long long h2=0,rh2=0;
int main()
{
    scanf("%s",s+1);
    long long ans = 0;
    int l = strlen(s+1);
    for(int i=1;i<=l;i++)
    {
        h1=h1*p1+s[i];
        rh1=s[i]*t1+rh1;
        t1=t1*p1;

        h2=h2*p2+s[i];
        rh2=s[i]*t2+rh2;
        t2=t2*p2;
        if(h1==rh1&&h2==rh2)
            dp[i]=dp[i/2]+1;
        ans+=dp[i];
    }
    cout<<ans<<endl;
}