xtu字符串 B. Power Strings

B. Power Strings

Time Limit: 3000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

 

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.

 

解题：求字符串的循环节长度。利用KMP的适配数组。如果字符长度可以被（字符长度-fail[字符长度])整除，循环节这是这个商，否则循环节长度为1，即就是这个字符本身。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <climits>
 #include <algorithm>
 #include <cmath>
 #define LL long long
 #define INF 0x3f3f3f
 using namespace std;
 const int maxn = ;
 char str[maxn];
 int fail[maxn];
 void getFail(int &len) {
     int i,j;
     len = strlen(str);
     fail[] = fail[];
     for(i = ; i < len; i++) {
         j = fail[i];
         while(j && str[j] != str[i]) j = fail[j];
         fail[i+] = str[j] == str[i] ? j+:;
     }
 }
 int main() {
     int len;
     while(gets(str) && str[] != '.') {
         getFail(len);
         if(len%(len-fail[len])) puts("");
         else printf("%d\n",len/(len-fail[len]));
     }
     return ;
 }