PKU 2406:Power Strings 【KMP】

Power Strings

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 70   Accepted Submission(s) : 27

Problem 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

abcdaaaaababab.

 

Sample Output

143

 

题意：给一个字符串S长度不超过10^6，求最大的n使得S由n个相同的字符串a连接而成，如："ababab"则由n=3个"ab"连接而成，"aaaa"由n=4个"a"连接而成，"abcd"则由n=1个"abcd"连接而成。

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。

这个是我在一篇博客上看到的。。。不懂为什么。。。。如果有大神知道。。。望不吝赐教！！！

AC-code:

#include<cstdio>#include<cstring>const int N=1001000;char str[N];int p[N],len;void getp(){int i=0,j=-1;p[i]=j;while(i<len){if(j==-1||str[i]==str[j]){i++,j++;if(str[i]==str[j])p[i]=p[j];elsep[i]=j;}elsej=p[j];}}int main(){while(scanf("%s",str)){if(!strcmp(str,"."))break;len=strlen(str);getp();if(len%(len-p[len])==0)printf("%d\n",len/(len-p[len]));elseprintf("1\n");}return 0;}