POJ2752 seek the Name, seek the Fame

Seek the Name, seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14203		Accepted: 7077

Description

The little cat is so famous, which many couples tramp over Hill and Dale to Byteland, and asked the little cat to give name s 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 a 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 isn't 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 is {' A ', ' ala ', ' Alala '}. Given the string S, could the little cat to write a program to calculate the length of possible prefix-suffix str Ings 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, 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 an 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

Test instructions: Find all substrings with the same prefix and suffix in a string

#include <iostream> #include <stdio.h> #include <string> #include <cstring> #include <cmath >using namespace Std;char s[1000009];int nt[1000009];int len;void getnext () {    int k=-1,j=0;    Nt[0]=-1;    while (J<len)    {        if (k<0 | | s[j]==s[k])        {            j + +;            k++;            nt[j]=k;        }        else            k=nt[k];}    } int Ans[400009];int Main () {while    (~scanf ("%s", s))    {        Len=strlen (s);        GetNext ();        for (int i=0;i<=len;i++)            cout<<nt[i]<< "";            cout<<endl;        int cnt=0;        Ans[0]=len;        for (int i=nt[len];i>0;i=nt[i])//next The nature of the array.            ans[++cnt]=i;        for (int i=cnt;i>0;i--)            cout<<ans[i]<< "";        cout<<ans[0]<<endl;    }    return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

POJ2752 seek the Name, seek the Fame