# poj 2406 Power Strings

 Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 41250 Accepted: 17150

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.


Sample Output

1
4
3


Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

本来想找后缀数组的题，发现可以用hash来O(n)解决。

 1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cmath>
5 #include<algorithm>
6 #include<cstring>
7 #include<queue>
8 #include<vector>
9 using namespace std;
10 typedef unsigned long long ULL;
11 const ULL BASE=127;
12 const int maxn=2000000;
13 char s[maxn];
14 int len;
15 ULL base[maxn],hash[maxn];
16 inline ULL get_hash(int l,int r){
17     return hash[r]-hash[l-1]*base[r-l+1];
18 }
19 int main(){
20     base[0]=1; for(int i=1;i<=1000000;i++) base[i]=base[i-1]*BASE;
21     for(;;){
22         scanf("%s",s+1); len=strlen(s+1);
23         if(s[1]=='.') break;
24         for(int i=1;i<=len;i++) hash[i]=hash[i-1]*BASE+s[i]-'a'+1;
25         bool vis=false;
26         for(int i=len-1;i>=1;i--){
27             if(get_hash(1,i)==get_hash(len-i+1,len)&&len%(len-i)==0){
28                 vis=true;
29                 printf("%d\n",len/(len-i));
30                 break;
31             }
32         }
33         if(vis==false) puts("1");
34     }
35     return 0;
36 }

posted @ 2016-04-19 20:03  CXCXCXC  阅读(187)  评论(0编辑  收藏