题目链接：http://poj.org/problem?id=2406
Time Limit: 3000MS Memory Limit: 65536K
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
题意：
求给出字符串，最多是多少个循环节组成的。
题解：
利用len % (len-Next[len]) == 0的话，len-Next[len]是最小循环节长度的性质。
AC代码：
#include<cstdio> #include<cstring> using namespace std; const int MAXpat = 1000000+5; char pat[MAXpat]; int Next[MAXpat],len; void getNext() { int i=0, j=-1; len=strlen(pat); Next[0]=-1; while(i<len) { if(j == -1 || pat[i] == pat[j]) Next[++i]=++j; else j=Next[j]; //printf("now i=%d j=%d next[%d]=%d pat[%d]=%c\n",i,j,i,Next[i],i,pat[i]); } } int main() { while(scanf("%s",pat)) { if(pat[0]==‘.‘ && pat[1]==‘\0‘) break; getNext(); if(len%(len-Next[len])==0) printf("%d\n",len/(len-Next[len])); else printf("1\n"); } }
注意：
①当且仅当len%(len-Next[len])==0时，len-Next[len]才是最小循环节长度。
②关于Next数组，我觉得这个图很不错的展示了Next数组存储了啥：
