poj2406PowerString（kmp求循环节）

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 28940 Accepted: 12080

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

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAXN = 1000005;char pattern[MAXN];int next[MAXN];void get_next(){    next[0]=next[1]=0;    int m = strlen(pattern);    for(int i=1; i<m; i++){        int j = next[i];        while(j && pattern[i]!=pattern[j]) j = next[j];        next[i+1] = (pattern[i]==pattern[j]? j+1 : 0);    }    int cir = m-next[m];//next[m] 最大为m-1，故cir最小为1，最大为m    if(m%cir==0) cout<<m/cir<<endl;    else cout<<1<<endl;}int main(){    while(scanf("%s",pattern) && pattern[0]!='.'){        get_next();    }    return 0;}

可打印字符要这样输入?

   while(gets(pattern)){      if(!strcmp(pattern , "."))