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POJ2406-Power Strings




Power Strings


Time Limit: 3000MS

 

Memory Limit: 65536K

Total Submissions: 55320

 

Accepted: 22983


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.


       题解:题意就是由一个字符串重复多遍组成,求最多重复次数。

       也就是求这个字符串的最小长度即可。很明显与KMP的fail数组有关,求出可以发现如果重复次数大于1,那么n-fail[n]必定是最小的字符串长度。随便列几次就可以发现了。

Code:

#include<cstdio>  
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1000005
int next[N],len;
int main()
{
char a[N];
while(~scanf("%s",a))
{
if(a[0]=='.')break;
len=strlen(a);
int i=0,j=-1;
next[0]=-1;
while(i<len)
{
if(j==-1||a[i]==a[j])
{
i++,j++;
next[i]=j;
}else j=next[j];
}
int k=len-next[len];
if(len%k==0)printf("%d\n",len/k);
else printf("1\n");
}
return 0;
}

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