Seek the Name, Seek the Fame POJ 2752【KMP next数组的应用】

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm: 

Step1. Connect the father's name and the mother's name, to a new string S. 
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S). 

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:) 

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above. 

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000. 

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

题意：给出一个字符串str，求出str中存在多少子串，使得这些子串既是str的前缀，又是str的后缀。从小到大依次输出这些子串的长度。

next数组就是存就是前一位最大长度值，即是str的前缀又是后缀的最大长度

计算某个字符对应的next值，就是看这个字符之前的字符串中有多大长度的相同前缀后缀

next数组就是存前一位最大前缀后缀长度值，（这个字符之前的字符串中有多大长度的相同前缀后缀）

一个数组的第len位前面的字符串刚好是输入的那个串，题目就是要求子串既是前缀又是后缀的长度，len位置存的是最大的，其他长度在前面，这就是Next数组神奇之处，我也不知道为什么！！

#include<cstdio>
#include<cstring>
char str[400400];
int next[400401];
int len;
void Get_Next()
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<len)
	{
		if(j==-1 || str[i]==str[j])
		{
			++i,++j;
			next[i]=j;
		}
		else
			j=next[j];
	}
}

void Print_Result(int i)
{
	
	if(next[i]!=-1)
	{
		Print_Result(next[i]);
		printf("%d ",i);
	}
	
}
int main()
{
	while(~scanf("%s",str))
	{
		len=strlen(str);
		Get_Next();
				
		Print_Result(next[len]);
		
		printf("%d\n",len);
	}
	
return 0;
}