http://codeforces.com/problemset/problem/7/D

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 exceed5·106. The string is case-sensitive.

Output

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

Sample test(s) input

a2A

output

1

input

abacaba

output

6

题目大意：一个字符串是回文串有一定的水平值他的水平值：如果他的前一半串和后一半串是k-1水平，那么他就是k。

解题思路：用kmp，具体就是把原串当作匹配串，倒过来匹配，当匹配的长度，大于等于当前的下标，这说明出现回文了

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int N=5000006;
int next[N];
char a[N];
bool is[N];
int i,n,j,ans;
int count(int r)//返回的是以r为结尾的字符串水平值是多少
{
    if(!r)
        return 1;
    int ret=1;
    while(r>0)
    {
        r=(r-1)>>1;
        if(is[r])
            ++ret;
        else
            break;
    }
    return ret;
}
int main()
{
    while(~scanf("%s",a))
    {
        memset(is,0,sizeof(is));
        n=strlen(a);
        next[0]=j=-1;
        for(i=1;i<n;i++)//KMP算法求next数组
        {
            while(j>-1&&a[j+1]!=a[i])
                j=next[j];
            if(a[j+1]==a[i])
                ++j;
            next[i]=j;
        }
        j=-1;
        for(i=n-1;i>=0;i--)//以原字符串为标准串于本身倒序匹配，当匹配结果大于等于一半是说明回文
        {
            while(j>-1&&a[j+1]!=a[i])
                j=next[j];
            if(a[i]==a[j+1])++j;
            while(j>=i)
            {
                is[i+j]=1;
                j=next[j];
            }
        }
        for(ans=i=0;i<n;++i)
            if(is[i])
                ans+=count(i);
        printf("%d\n",ans);
    }
    return 0;
}