Poj--2752seek the Name, seek the Fame (KMP to find the same string before and after)

Seek the Name, seek the Fame
Time limit:2000ms Memory limit:65536k
Total submissions:16035 accepted:8158
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


Ababcababababcabab
Aaaaa
Sample Output


2 4 9 18
1 2 3) 4 5
Source

POJ Monthly--2006.01.22,zeyuan Zhu

Given a string s, the length of a substring that is both a prefix and a suffix from small to large output s.
This problem is very simple, borrow the next array of the KMP algorithm, set S length is N,
The S-string itself must satisfy the condition. Other substrings that satisfy the condition have a feature,
The last character of the substring must be the same as the last character of S.
This is when the next array works.
From the n-1 bit both the last one starts rolling back, if s[next[n-1]] = = S[n-1],
The substring s[0,1,2,..., next[n-1]] is a substring that satisfies the condition.
Then Judge s[next[next[n-1]] [= S[n-1] is set up, so keep rolling back,
Until next[next[.....next[n-1]] = =-1. Save the answer from the big to the small, then from small to large output can.

#include <iostream> #include <cstdio> #include <cstring>using namespace Std;int next[400005];char str [400005];int ans[400005];int cnt;int len;void getNext () {next[0] = -1;int i = 0, j = -1;while (i < len) {if (j = =-1 | | s Tr[i] = = Str[j]) {++i;++j; Next[i] = j;} else J = Next[j];}} int main () {while (scanf ("%s", str)! = EOF) {len = strlen (str); GetNext (); cnt = 0;int T = next[len-1];while (t! =-1) {if (s Tr[t] = = str[len-1]) ans[cnt++] = T+1;t = next[t];//jumps back to the previous position matching the last character}for (int i = cnt-1; I >= 0; i) {printf ("%d ", Ans[i]);} printf ("%d\n", Len);} return 0;}

Poj--2752seek the Name, seek the Fame (KMP to find the same string before and after)