POJ 2406 Power Strings KMP算法

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分类专栏： ACM-字符串-KMP 文章标签： poj 算法 kmp

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 49787		Accepted: 20747

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

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Link: http://poj.org/problem?id=2406

这里给出一个定理：

假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]

AC Code：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000000+50;
int f [maxn];
int ls,la;
void getfill(char* s)
{
    memset(f,0,sizeof(f));
    for(int i=1;i<ls;i++)
    {
        int j=f[i];
        while(j && s[i]!=s[j])
            j=f[j];
        f[i+1]=(s[i]==s[j])?j+1:0;
    }
}
int find(char* a,char* s)
{
    int ans=0;
    getfill(s);int j=0;
    for(int i=0;i<la;i++)
    {
        while(j && a[i]!=s[j])
            j=f[j];
        if(a[i]==s[j])
            j++;
        if(j==ls){
            ans++;
        }
    }
    return ans;
}
int main()
{
    char str[maxn];
    while(~scanf("%s",str)){
        if(str[0]=='.') break;
        ls=strlen(str);
        getfill(str);
        if(ls%(ls-f[ls])==0) printf("%d\n",ls/(ls-f[ls]));
        else printf("1\n");
    }
    return 0;
}