POJ 2406 Power Strings (KMP)

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: }