Power Strings
Time Limit: 3000MS
Memory Limit: 65536K
Total Submissions: 29663
Accepted: 12387
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
::初学KMP真心不容易，理解起来很吃力，还好今天老师有讲解KMP的基本原理，现在思路清晰了一点，赶紧做这道以前暴力过的题。
证明：（PKU 2406 POWER STRINGS --- 字符串匹配，KMP算法）
定理：假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]
例子证明：
设S=q1q2q3q4q5q6q7q8，并设next[8] = 6，此时str = S[len - next[len]] = q1q2，由字符串特征向量next的定义可知，q1q2q3q4q5q6 = q3q4q5q6q7q8，即有q1q2＝q3q4，q3q4＝q5q6，q5q6＝q7q8，即q1q2为循环子串，且易知为最短循环子串。由以上过程可知，若len可以被len - next[len]整除，则S存在循环子串，否则不存在。
解法：利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，则最大循环次数n为len/(len - next[len])，否则为1。
1: #include <cstdio>
2: #include <cstring>
3: #include <algorithm>
4: using namespace std;
5: const int maxn=1e6;
6: char s[maxn+10];
7: int next[maxn+10];
8:
9: void get_next(char s[],int len)
10: {
11: int i=0,j=-1;
12: next[0]=-1;
13: while(i<len)
14: {
15: if(j==-1||s[i]==s[j]) {j++; i++; next[i]=j;}
16: else j=next[j];
17: }
18: }
19:
20: int main()
21: {
22: while(scanf("%s",s),s[0]!='.')
23: {
24: int len=strlen(s);
25: get_next(s,len);
26: if(len%(len-next[len])==0)
27: printf("%d\n",len/(len-next[len]));
28: else
29: printf("1\n");
30: }
31: return 0;
32: }