poj2406 Power Strings

地址：http://poj.org/problem?id=2406

题目：

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 47529		Accepted: 19823

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.

Source

Waterloo local 2002.07.01

 

思路：kmp+最小循环节

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
const double eps=1e-8;
const int K=1e6+7;
const int mod=1e9+7;

int nt[K];
char sa[K],sb[K];
void kmp_next(char *T,int *next)
{
    next[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[i]!=T[j]) j=next[j-1];
        if(T[i]==T[j])  j++;
        next[i]=j;
    }
}
int kmp(char *S,char *T,int *next)
{
    int ans=0;
    int ls=strlen(S),lt=strlen(T);
    kmp_next(T,next);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j]) j=next[j-1];
        if(S[i]==T[j])  j++;
        if(j==lt)   ans++;
    }
    return ans;
}
int main(void)
{
    while(1)
    {
        scanf("%s",sa);
        if(sa[0]=='.')break;
        kmp_next(sa,nt);
        int len=strlen(sa);
        int ans=len-nt[len-1];
        printf("%d\n",len%ans?1:len/ans);
    }
    return 0;
}