POJ 2406 Power Strings(KMP next[]函数）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23112		Accepted: 9691

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

 

 

 

简单的KMP题目：

 

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

const int MAXN=1000000;
char str[MAXN];
int next[MAXN];
int len;
void getNext()
{
    int j,k;
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||str[j]==str[k])
           next[++j]=++k;
        else k=next[k];
    }
}
int main()
{
   // freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    while(scanf("%s",&str)!=EOF)
    {
        if(strcmp(str,".")==0) break;//这个地方要注意
        len=strlen(str);
        getNext();
        if(len%(len-next[len])==0&&len/(len-next[len])>1)printf("%d\n",len/(len-next[len]));
        else printf("1\n");
    }
    return 0;
}