POJ2406--Power Strings（KMP变形）

Do more with less

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.

题意

字符串abcd 是 abcd 的一次方

字符串aaaa 是 a 的四次方

字符串ababab 是 ab 的3次次方

输入 . 表示测试结束

思路

如果是一个没有重复的串，那它的next数组会一直是0

如果next数组从第二个数起一直增加，说明这个序列是由一个字母组成

如果倒序检索next数组，遇到的第一个0的时候，前面的序列一定不能用次方的形式表示

代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char p[1000005];
int nextp[1000005];
int m;
int getnextp()
{
int i = 0;
int j = -1;
nextp[0] = -1;
while(i <= m)
{
if(p[i] == p[j] || j == -1)
{
j++;
i++;
nextp[i] = j;
}
else
{
j = nextp[j];
}
}
i = m;
while(nextp[i] > nextp[i-1] && i >= 0)
{
i--;
}
if(i == 0)
return m;
if(nextp[i] == 0 && m % i == 0)
return m /i;
else
return 1;
}
int main()
{
while(scanf("%s",p) && p[0] != '.')
{
m = strlen(p);
printf("%d\n", getnextp());
}
return 0;
}