poj2406 Power Strings（kmp失配函数）

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 39291		Accepted: 16315

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。

       应用KMP算法中的失配函数，如果是一个周期串那么错位部分(n-f[n])一定是一个最小循环节（仔细想想f函数的意义），则答案为i/(n-f[n])。否则ans=1。

 　　时间复杂度为O(n)。

【代码】

 

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(a,b,c) for(int a=(b);a<=(c);a++)
using namespace std;

const int maxn = 1000000+10;

void getFail(char* P,int* f) {
    int m=strlen(P);
    f[0]=f[1]=0;
    for(int i=1;i<m;i++) {
        int j=f[i];
        while(j && P[i]!=P[j]) j=f[j];
        f[i+1]=P[i]==P[j]?j+1:0;
    }
}

char s[maxn];
int f[maxn];

int main() {
    while(scanf("%s",s)==1 && s[0]!='.') {
        getFail(s,f);
        int n=strlen(s);
        if(f[n]>0 && n%(n-f[n])==0) printf("%d\n",n/(n-f[n]));
        else printf("%d\n",1);
    }
    return 0;
}