poj 2406

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 26943 Accepted: 11281

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

Waterloo local 2002.07.01

主要是next函数，才发现n个字符是n+1个next值（next0-----nextn），然后l%(l-nextn)==0代表字符串由l/(l-nextn)个相同字符串组成。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;char s[1000005];int next[1000005];void getnext(){    int l=strlen(s);    int j=0,k=-1;    next[0]=-1;    while(j<l)    {        if(k==-1||s[j]==s[k])        {            j++;            k++;            next[j]=k;        }        else k=next[k];    }    int h=l-k;    if(l%h==0)cout<<l/h<<endl;    else cout<<1<<endl;}int main(){    while(scanf("%s",&s))    {        if(s[0]=='.')return 0;        getnext();    }}