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

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 ;

 }

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