Poj-2406-Power Strings【KMP】

传送门：http://poj.org/problem?id=2406常发生

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 50464 Accepted: 21048

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

题意：拿第三组数据来看 找到子串ab 则s为ababab  a为ab  n为s除以a   

第二组数据 匹配子串为一个a（这里就是next数组的运用）s为aaaa  a为a  n为4

#include<cstdio>#include<cstring>#define M 1000005char s[M];int next[M];int len;void Next(){int i=0;int j=-1;next[0]=-1;j=next[i];while(i<len){if(j==-1||s[i]==s[j]){next[++i]=++j;//i++;j++;//next[i]=j;}elsej=next[j];}}int main(){while(scanf("%s",s)!=EOF){if(s[0]=='.')break;len=strlen(s);   //不能两次 int  Next();if(len%(len-next[len])==0)   //len-next[len]  求公共子串长度printf("%d\n",len/(len-next[len]));elseprintf("1\n");}return 0;}