POJ2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 42797		Accepted: 17855

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：给一个字符串S长度不超过10^6，求最大的n使得S由n个相同的字符串a连接而成，如："ababab"则由n=3个"ab"连接而成，"aaaa"由n=4个"a"连接而成，"abcd"则由n=1个"abcd"连接而成。

定理：假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]

例子证明：
设S=q1q2q3q4q5q6q7q8，并设next[8]
= 6，此时str = S[len - next[len]] = q1q2，由字符串特征向量next的定义可知，q1q2q3q4q5q6 =
q3q4q5q6q7q8，即有q1q2＝q3q4，q3q4＝q5q6，q5q6＝q7q8，即q1q2为循环子串，且易知为最短循环子串。由以上过程可知，若len可以被len
- next[len]整除，则S存在循环子串，否则不存在。

解法：利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，则最大循环次数n为len/(len - next[len])，否则为1。

注意：1.while多输入时一定要初始化 2.abbabbaa串 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char p[1000005];
int next[1000005];
void nxt(char *p)
{
int plen=strlen(p);
next[0]=-1;
int j=0,k=-1;
while(j<plen)
{
if(k==-1 || p[k]==p[j])
{
next[j+1]=k+1;
j++;k++;
}
else k=next[k];
}
}

int main()
{
while(~scanf("%s",&p) && p[0]!='.')
{
memset(next,0,sizeof next);//
nxt(p);
int len=strlen(p);
//	cout<<next[len]<<endl;
if(len%(len-next[len])==0)
printf("%d\n",len/(len-next[len]));
else printf("1\n");
}

}