poj2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 52620		Accepted: 21917

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

大致题意：求一个字符串的最小循环节.

分析：

      KMP算法的经典应用.利用next数组来求,next[i]表示的是匹配到第i位后要跳回到第next[i]位去继续匹配，这里的匹配可以相当于是自己和自己匹配.(next[i],i]表示的字符串就可能是一个循环节.如果i - next[i] | len,那么i - next[i]就是答案，否则循环节就是字符串本身.因为每次跳到next[i]的时候，(next[i],i]的字符串已经和前面的匹配好了，而(2*next[i] - i,next[i]]也与前面匹配好了，所以(next[i],i]就有可能是一个循环节.如果不整除的话，就会有余数位不能被匹配，也就不能构成循环节了.

      一开始的想法是求min{i - next[i]},事实上是错的，因为只保证了i前面的能匹配上，后面的不能保证.为了保证这个循环节是整个字符串的循环节，i取len.

strlen函数非常慢，要少用QAQ

#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

char s[1000010];
int nextt[1000010];

int main()
{
    while (scanf("%s", s + 1) != 0)
    {
        if (s[1] == '.')
            break;
        nextt[0] = 0;
        int j = 0, len = strlen(s + 1);
        for (int i = 2; i <= len; i++)
        {
            while (j && s[j + 1] != s[i])
                j = nextt[j];
            if (s[j + 1] == s[i])
                j++;
            nextt[i] = j;
        }
        int t = len - nextt[len];
        if (len % t == 0)
            printf("%d
", len / t);
        else
            printf("%d
", 1);
    }

    return 0;
}