Codeforces 7D Palindrome Degree 字符串hash DP

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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 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 test(s)

Input

a2A

Output

1

Input

abacaba

Output

6

题意：如果一个字符串s是长度为n的k度回文，它的长度为的前缀和后缀为k-1度回文。字符串的回文度，就是这个k的最大值。

给出一个字符串，求所以前缀的回文度的和。

思路：假设当前前缀的尾端为第i个字符，维护这个前缀字符串的正向hash值和逆向hash值，如果两个值相等的话（就是回文啦），它的度就是这个前缀一半长度的字符串的度+1，即dp[i]=dp[i>>1]+1;

代码：

#include<cstdio>
#include<cstring>
#define uLL unsigned long long
#define maxn 5000005
char s[maxn];
int dp[maxn];
int main()
{
	scanf("%s", s);
	memset(dp, 0, sizeof(dp));
	dp[0] = 1;
	uLL ha = s[0];
	uLL haf = s[0];
	uLL hb = 1;
	int ans = 1;
	for (int i = 1; s[i]; i++)
	{
		hb *= 131;
		ha = ha * 131 + s[i];
		haf = s[i] * hb + haf;
		if (ha == haf) dp[i] = dp[(i - 1) >> 1] + 1;
		ans += dp[i];
	}
	printf("%d\n", ans);
	return 0;
}