POJ 2752 KMP求前缀后缀的最大匹配长度

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

参考：https://blog.csdn.net/guhaiteng/article/details/52108690

next[i]的意义就是：前面长度为i的字串的【前缀和后缀的最大匹配长度】

以ababcababababcabab为例：

字符串的总长度为m=18,18是最大的前后缀匹配长度，m中的前缀和后缀长为next[m]=9;

此时自然是在前缀蓝，和后缀红中找比9更小的前后缀和；

注意了，红和蓝是一样的，只需要在蓝中找其前后缀长度即可next[9]=4;

依次类推，next[2]=0;则下标2之前的没有能匹配的了。

代码如下： 

#include
#include
#include
#include
#include
#include
#define N 10000
#define M 1000010
using namespace std;
int nex[M];
char b[M];
int n,m;
void nexx(){
    int k=-1,j=0;
    nex[0]=-1;
    while(j