poj2406-Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 27230		Accepted: 11401

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.

 

 

地址：http://poj.org/problem?id=2406

题意：ababab，ab重复了三次，于是输出3，abcd只有自身重复了一次，输出1.。。。。。

这么一道题做了一小天，一直在想，终于想出一个自我感觉很良好的方法，把输入的字符串复制一下，还是以a，b的形式，把b后移，每次移动j-next[j]个位子，然后把后面多出去的数改成'\0'，于是用这个方式实现了好久，最终在一个细节上以失败告终，不死心想看看别人都怎么做的，看到了一句话解释next函数，当一个字符串以0为起始下标时，next[i]可以描述为"不为自身的最大首尾重复子串长度"。原来只会求next，却没去想它真正代表的含义是什么，这样就好想了，如果是abcabcabc的话，它的next为-1，-1，-1,0,1,2,3,4,5，可以发现，如果是有重复的子段，那么它的next一定是这种情况，如果len%（len-next[len-1]-1)==0的话，那么len/（len-next[len-1]-1)一定是要求的那个值。。。。。。。。知识没学透的代价啊。。。。。。。。

 

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
char B[1000100];
int next[1000010];

void prekmp() {
    next[0] = -1;
    int j = -1;
    for (int i = 1; B[i]; i++) {
        while (j != -1 && B[i] != B[j + 1])
            j = next[j];
        if (B[i] == B[j + 1]) j++;
        next[i] = j;
    }
    int x=strlen(B);
    if(x%(x-next[x-1]-1)==0)printf("%d\n",x/(x-next[x-1]-1));
    else printf("1\n");
}


int main()
{
    while(scanf("%s",B)&&strcmp(B,"."))
    {
        prekmp();
    }
    return 0;
}