Codeforces Beta Round #7 D. Palindrome Degree

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

哈希判断回文。。

/* *===================== *File Name:a.cpp *Author: qqspeed *Date: 2014年 07月 19日 星期六 10:18:00 CST *===================== */#include <stdio.h>#include <string.h>#include <iostream>#include <string>#include <queue>#include <stack>#include <map>#include <vector>#include <stdlib.h>#include <algorithm>using namespace std;typedef long long ll;#define rep(i,s,t) for(int i=s;i<t;i++)#define red(i,s,t) for(int i=s-1;i>=t;i--)#define ree(i,now) for(int i=head[now];i!=-1;i=edge[i].next)#define clr(a,v) memset(a,v,sizeof a)#define L t<<1#define R t<<1|1#define MID int mid=(l+r)>>1#define max(a,b) (a<b?b:a)#define min(a,b) (a<b?a:b)#define SQR(a) ((a)*(a))inline int input(){int ret=0;bool isN=0;char c=getchar();while(c<'0' || c>'9'){if(c=='-') isN=1;c=getchar();}while(c>='0' && c<='9'){ret=ret*10+c-'0';c=getchar();}return isN?-ret:ret;}inline void output(ll x){        if(x<0){            putchar('-');x=-x;        }        int len=0,data[20];        while(x){            data[len++]=x%10;x/=10;        }        if(!len)    data[len++]=0;        while(len--)           putchar(data[len]+48);        putchar('\n');  }  const ll mod=1000000007;const ll base=137;const int N=5000005;char a[N],b[N];int vis[N];ll p[N],h[N],rh[N];ll tmp[N];bool inq[N];int l;inline void init(){p[0]=1;rep(i,1,N) p[i]=p[i-1]*base%mod;rep(i,0,l) b[l-i-1]=a[i];}inline void hash(char *s,ll *ha){ha[0]=s[0];rep(i,1,l){ha[i]=ha[i-1]+s[i]*p[i];ha[i]%=mod;}}inline ll get(int s,int t,ll *h){if(s==0) return h[t];else return (h[t]-h[s-1]+mod)%mod;}inline bool is_Palidrome(int a,int b,int c,int d){ll h1=get(a,b,h);ll h2=get(l-d-1,l-c-1,rh);return h1*p[l-d-1]%mod == h2*p[a]%mod;}inline ll solve(){ll ans=1;tmp[0]=1;rep(i,1,l){int a=0;int b=(i+1)/2-1;int c=(i%2==1?b+1:b+2);int d=i;if(is_Palidrome(a,b,c,d)){tmp[i]=tmp[b]+1;ans+=tmp[i];//printf("%d %d\n",i,tmp[i]);}}return ans;}int main(){scanf("%s",a);l=strlen(a);init();clr(vis,0);hash(a,h),hash(b,rh);output(solve());return 0;}