Poj2752 seek the name, seek the fame (use of the next array)

Question link: http://poj.org/problem? Id = 2752

Seek the name, seek the fame

Time limit:2000 ms		Memory limit:65536 K
Total submissions:12018		Accepted:5906

Description

The little cat is so famous, that could 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, cocould 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

Source

Poj monthly -- 2006.01.22, zeyuan Zhu

Search for substrings with the same prefix and suffix.

The Code is as follows:

# Include <cstdio> # include <cstring> # include <algorithm> # include <iostream> using namespace STD; # define maxn limit 17int next [maxn]; int ans [maxn]; int Len; void getnext (char T []) {int I = 0, j =-1; next [0] =-1; while (I <Len) {If (j =-1 | T [I] = T [J]) {I ++, J ++; next [I] = J ;} else J = next [J] ;}} int main () {char ss [maxn]; int length; while (~ Scanf ("% s", SS) {Len = strlen (SS); getnext (SS); int n = 0; int I = Len; ans [0] = Len; while (next [I]> 0) // scan the next array upside down {// Recursively search for substrings with the same pre-and post-substrings I = next [I]; ans [++ N] = I ;}for (I = N; I> = 0; I --) printf ("% d", ANS [I]); printf ("\ n");} return 0 ;}