Seek the Name, Seek the Fame next数组的应用

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

ababcababababcababaaaaa

Sample Output

2 4 9 181 2 3 4 5

题目大意：求一个字符串前后缀相等的长度。很明显，最大长度是next[len]，然后不断的取next就得到了全部前后缀相等的长度。在运行时发现我的next数组含义有点不一样，因为我比较的是s[j + 1] == s[i]，所以next[j]是坐标，加一后才是长度

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;const int N = 4e5 + 3;char s[N];int next[N];void GetNext(){    int j = -1, len = strlen(s);    next[0] = -1;    for(int i = 1; i < len; i++)    {        while(j != -1 && s[j + 1] != s[i])            j = next[j];        if(s[j + 1] == s[i]) j++;        next[i] = j;    }}int main(){    while(~scanf("%s", s))    {        GetNext();        int i = strlen(s) - 1, k = 0, m[N];        while(next[i] != -1)        {            m[k++] = next[i];            i = next[i];        }        for(i = k - 1; i >= 0; i--)            printf("%d ", m[i] + 1);        printf("%d\n", strlen(s));    }    return 0;}