Power Strings

Power Strings

Time Limit: 1000MS    Memory limit: 65536K

题目描述

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).

输入

 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.

输出

 For each s you should print the largest n such that s = a^n for some string a.

示例输入

abcdaaaaababab.

示例输出

143

提示

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

在运行整段字符串的的时候最后的的时候i-j就是字符串的最小子字符串

 

#include<stdio.h>#include<string.h>char t[110000000];int next[11000000];int main(){    int n,m,i,j;    int len1;    while(~scanf("%s",t)&&strcmp(t,".")!=0)    {        len1=strlen(t);        i=0,j=-1;        next[0]=-1;        while(i<len1)        {            if(j==-1||t[i]==t[j])            {                ++i;                ++j;                if(t[i]!=t[j]) next[i]=j;                else next[i]=next[j];            }            else j=next[j];        }        if(len1%(len1-j)==0)        {            int l=len1/(len1-j);            printf("%d\n",l);        }        else            printf("1\n");    }    return 0;}