POJ 2406 Power Strings

这道题目的意思是如果一个字符串可以由若干个连续的子串组成，比如aaaa是有四个a组成所以是a^4，所以输出4。但是aasaasaa应该输出1因为aas这个循环节循环不到最后、、、所以解题思路就很清晰了啊、、

就是找到最后一个的循环节，p = len-next[len],如果len%p==0，说明是循环节。否则这不是整个的循环节、输出1。

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 28388 Accepted: 11867

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 <stdio.h>#include <string.h>#include <cmath>#include <iostream>#include <string>using namespace std;int next[1010000];string str;int main(){    int len, i, j;    while(cin >>str)    {        len = str.length();        if(len == 1 && str[0] == '.')            break;        next[0] = 0;        next[1] = 0;        for(i = 1; i < len; i++)        {            j = next[i];            while(j && str[i]!= str[j])                j = next[j];            if(str[i] == str[j])                next[i+1] = j+1;            else                next[i+1] = 0;        }        if(len%(len-next[len]))            cout <<'1'<<endl;        else            cout <<len/(len-next[len])<<endl;    }    return 0;}