Power Strings 【poj-2406】【KMP】

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 50546 Accepted: 21082

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

题解：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-1]整除，最短循环子串为S[len - next[len-1]]

根据算法的不同，这里的next[len-1]代表整个字符串(即算上最后一位字符)的最大匹配值。

若不能整除，输出1；

代码如下：

#include<cstdio>#include<cstring>char P[1000005];int Next[1000005];int main(){while(~scanf("%s",P)){if(P[0]=='.') break;int i,k=0;memset(Next,0,sizeof(Next));int n=strlen(P);Next[0]=0;for(i=1;i<n;i++){while(k>0&&P[i]!=P[k])k=Next[k-1];if(P[i]==P[k])k++;Next[i]=k;}if(n%(n-Next[n-1])==0)printf("%d\n",n/(n-Next[n-1]));elseprintf("1\n");}return 0;}