poj 2406 Power Strings（KMP）

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 44585 Accepted: 18625

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.

思路：KMP，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。所以思路和上面一样，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

#include<cstdio>#include<cstring>int next[1000010];char str[1000010];void get_next(){    int i = 0 , j = -1;    next[0] = -1;    int len=strlen(str);    while(i < len)    {        if(j == -1 || str[i] == str[j]){            ++i; ++j;            next[i] = j;        }        else            j = next[j];    }}int main(){    while(~scanf("%s",str))     {        if(str[0] == '.')            break;        int len = strlen(str);        get_next();        if(len%(len-next[len]) == 0)            printf("%d\n",len / (len-next[len]));        else            printf("1\n");    }    return 0;}