Codeforces Beta Round #7--D. Palindrome Degree（hash）

D. Palindrome Degree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

 

哈希一下，从左向右和从右向左的哈希值是一样的话，那么该串是回文串，所以求出从开头到第i个字符是不是回文串

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
#define Mod (LL)(1e9+7)
char str[6000000] ;
int dp[6000000] ;
int main()
{
    LL ans , i , l , p , q , temp ;
    while( gets(str) != 0 ) {
        memset(dp,0,sizeof(dp)) ;
        l = strlen(str) ;
        if( l == 0 ) {
            printf("0\n") ;
            continue ;
        }
        dp[0] = 1 ;
        ans = 1 ;
        p = q = str[0] ;
        temp = 1 ;
        for(i = 1 ; i < l ; i++) {
            p = (p*131+str[i]) % Mod ;
            temp = (temp*131) % Mod ;
            q = (str[i]*temp + q) % Mod ;
            if( p == q ) {
                dp[i] = dp[(i-1)/2] + 1 ;
                ans += dp[i] ;
            }
        }
        printf("%I64d\n", ans) ;
    }
    return 0 ;
}