   Online JudgeProblem SetAuthorsOnline ContestsUser
Web Board
F.A.Qs
Statistical Charts
Problems
Submit Problem
Online Status
Prob.ID:
Register
Authors ranklist
Current Contest
Past Contests
Scheduled Contests
Award Contest
Register

## 水平不够，，，，w了特别久

Posted by leiyongZL at 2016-07-25 16:14:48 on Problem 2406
```/******

Submit

Status

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.

+
********/
#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int next;
void getnext(char *S,int& R)
{
int j=0;
next=0;
for(int i=1; S[i]!='\0'; i++)
{

while(j>0&&S[i]!=S[j])
{
j=next[j-1];
}
if(S[i]==S[j])
{
j++;
}
next[i]=j;
}
}
int main()
{
char S;
while(scanf("%s",S)&&S!='.')
{
memset(next,0,sizeof(next));
int R=0;
getnext(S,R);
int l=strlen(S);
int j=next[l-1];
int m=l-j,k=j,B;
while(j>0)
{
if(j==m)
{
B=1;
break;
}
j=next[j-1];
int n=k-j;
if(n!=m)
{
B=0;
break;
}
k=j;
}
//cout<<m<<endl;
if(B)
cout<<l/(m)<<endl;
else
cout<<"1"<<endl;
}
return 0;
}
```

Followed by: Home Page Go Back To top