KMP算法：next数组变形

Seek the Name, Seek the Fame

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

给一个字符串，求所有前缀等于后缀的情况，得到他们的字符长度。

KMP求出next数组，每次去掉next[i]到i的一段字符，剩余字串仍满足条件，直到找到头为止。

next[i]！=0的，都是模式串的前缀和后缀相同的字符数。

从最后一个字符找，一直到第一个为止。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int Next[400005];int len;char s[400005];int ans[400005];void getNext(){int j=0,k=-1;Next[0]=-1;while(j<len){if(k==-1||s[j]==s[k])Next[++j]=++k;elsek=Next[k];}}int main(){while(cin>>s){int cnt=0;len=strlen(s);getNext();int num=len;ans[cnt++]=len;while(Next[num]!=0){num=Next[num];ans[cnt++]=num;}for(int i=cnt-1;i>0;i--){cout<<ans[i]<<" ";}cout<<ans[0]<<endl;}return 0;}