Poj 2752 seek the name, seek the fame KMP

Language:DefaultSeek the name, seek the fame Time limit:2000 ms   Memory limit:65536 K Total submissions:12466   Accepted:6139 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

Give a string 'str' and find all the lengths of both the prefix and suffix. This is the application of the next array of KMP.

Code:

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include <string> # include <map> # include <stack> # include <vector> # include <set> # include <queue> # pragma comment (linker, "/Stack: 102400000,102400000") # define maxn 400010 # define maxn 2005 # define mod 1000000009 # define INF 0x3f3f3f # define PI ACOs (-1.0) # define EPS 1e-6typedef long ll; using namespace STD; int ne XT [maxn], ANS [maxn]; char STR [maxn]; void get_next () {int Len = strlen (STR); int I = 0, j =-1; next [0] =-1; while (I <Len) {If (j =-1 | STR [I] = STR [J]) {I ++; j ++; next [I] = J;} else J = next [J];} ans [0] = Len; int T = 1; j = Len; while (next [J]> 0) // roll out all the lengths and record them in the ANS Array {ans [T] = next [J]; t ++; j = next [J];} For (INT I = T-1; I> 0; I --) printf ("% d", ANS [I]); printf ("% d \ n", ANS [0]);} int main () {While (scanf ("% s", STR )! = EOF) get_next (); Return 0;}/* ababcabababcababaaaaa */

Poj 2752 seek the name, seek the fame KMP