题目链接：https://vjudge.net/problem/POJ-2406
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 52631 | Accepted: 21921 |
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
KMP：
求字符串的最小循环节。
1) 如果字符串长度能被“初步的最小循环节”整除，那么这个就是他的最小循环节。
2) 如果字符串长度不能被“初步的最小循环节”整除，那么这个只是它通过补全后的最小循环节。而它实际的最小循环节为自己。
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 typedef long long LL; 14 const double eps = 1e-6; 15 const int INF = 2e9; 16 const LL LNF = 9e18; 17 const int MOD = 1e9+7; 18 const int MAXN = 2e6+10; 19 20 char x[MAXN]; 21 int Next[MAXN]; 22 23 void get_next(char x[], int m) 24 { 25 int i, j; 26 j = Next[0] = -1; 27 i = 0; 28 while(i<m) 29 { 30 while(j!=-1 && x[i]!=x[j]) j = Next[j]; 31 Next[++i] = ++j; 32 } 33 } 34 35 int main() 36 { 37 while(scanf("%s", x) && strcmp(x, ".")) 38 { 39 int len = strlen(x); 40 get_next(x, len); 41 int r = len-Next[len]; //求出最小循环节 42 if(len%r) //如果长度/最小循环节除不尽，则对于此字符串来说，实际的最小循环节是自己 43 printf("1 "); 44 else 45 printf("%d ", len/r); 46 } 47 }
后缀数组：
