poj 2406 Power Strings

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

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

题意：自己看（最后遇到“.”要break）

解：

len为整个字符串的长度

想一想next[i]的含义

每次匹配失败后，都要回到next[i]开始重新匹配

也就是说  next[i]+1 -> len  这段字符 (令k为这段字符的长度)和

1 -> 1+k  这段字符 是相同的

OK，如果(len%(len-next[len])==0)  ==>  存在 len/(len-next[len]) 段 解

否则只有1段

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<map>
using namespace std;
const int N=1e7;
char s[N];
int len,ne[N],j;
int main()
{
    while(scanf("%s",s+1)!=EOF)
    {
        if(s[1]=='.') break;
        len=strlen(s+1);
        for(int i=0;i<=len;++i) ne[i]=0;
        for(int i=2;i<=len;++i)
        {
            j=ne[i-1];
            while(j && s[i]!=s[j+1]) j=ne[j];
            if(s[i]==s[j+1]) ne[i]=j+1;
        }
        if(len%(len-ne[len])==0) printf("%d
",len/(len-ne[len]));
        else printf("1
");
    }
    return 0;
}