POJ 2406 Power Strings（最小循环节的应用）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 27359		Accepted: 11450

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

                题目大意：给你一个串s，求a^n=s，求出最大的n.记得以前曾经暴力解过此类题目。

 

        题目地址：Power Strings

由于是a^n跟剪花布条类似，不能重合。

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
char a[1000005];
int len,next[1000005];

void getnext()
{
    int i,j;
    next[0]=0,next[1]=0;
    for(i=1;i<len;i++)
    {
         j=next[i];
         while(j&&a[i]!=a[j])
           j=next[j];
         if(a[i]==a[j])
            next[i+1]=j+1;
          else
            next[i+1]=0;
    }
}

int main()
{
     while(scanf("%s",a))
     {
          if(strcmp(".",a)==0)
               break;
          len=strlen(a);
          getnext();

          int res;
          if(len%(len-next[len]))//len-next[len]为最小循环节
             res=1;   //需要排除AZAZAZA的情况，此时应该输出1而不是3.考虑重合或者不重合
            else
               res=len/(len-next[len]);
          printf("%d\n",res);
     }
     return 0;
}