Problem 1011 Power Strings
Accept: 914 Submit: 2751
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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).

Each test case is a line of input representing s, a string of printable characters. For each s you should print the largest n such that s = a^n for some string a. 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.

Sample Input

abcd
aaaa
ababab
.
Sample Output

1
4
3

本題為簡單的KMP匹配演算法，，是我在KMP這個演算法中做的第一道題.
下面是我的AC程式碼

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char str[1000005];
int next[1000005];

int kmp()
{
    int i=0,j=-1;
    next[0]=-1;
    while
 
(str[i])
    {
        if(j==-1||str[i]==str[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else j=next[j];
    }
    return j;
}

int main()
{
    while(gets(str)&&str[0]!='.')
    {

        int len=strlen(str);
        int r=len-kmp();
        if(len%r==0)
        {
            printf("%d\n",len/r);
        }
        else printf("1\n");
    }
    return 0;
}