(KMP 1.5) HDU 1358 Period (use next array to find the smallest loop-the number of cyclic sections for the I-character)

Topic:

PeriodTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3813 Accepted Submission (s): 1862


Problem descriptionfor Each prefix of a given string S with N characters (each character have an ASCII code between 126, inclusive), we want to know whether the prefix is a periodic string. That's, for each I (2 <= i <= N) we want to know the largest K > 1 (if there are one) such that the prefix of S W ith length I can be written as AK, which is A concatenated K times, for some string A. Of course, we also want to know the period K.

Inputthe input file consists of several test cases. Each test case consists of lines. The first one contains n (2 <= n <= 1 000 000), haven size of the string S. The second line contains the string S. The input file ends with a line, has the number zero on it.

Outputfor each test case, output ' test Case # ' and the consecutive test case number on a '; Then, for each prefix with length I that have a period K > 1, output the prefix size I and the period K separated by a s Ingle Space; The prefix sizes must is in increasing order. Print a blank line after each test case.

Sample Input

3aaa12aabaabaabaab0

Sample Output

Test Case #12 3Test case #22 26 29 312 4

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Topic Analysis:

Kmp. Simple question. This question is in fact the same idea as the KMP 1.4.

The code is as follows:

/* * hdu1358.cpp * *  Created on:2015 April 18 *      author:administrator * * #include <iostream> #include < algorithm> #include <cstdio> #include <cstring>using namespace std;const int maxn = 1000001;int m;// The length of the target string char pattern[maxn];//pattern string int nnext[maxn];//next array. Direct from next may be the same as the predetermined name in the system./*o (m) time for next array */void Get_next () {m =  Strlen (pattern); nnext[0] = nnext[1] = 0;for (int i = 1; i < m; i++) {int J = nnext[i];while (J && pattern[i]! = PATTERN[J]) j = nnext[j];nnext[i + 1] = pattern[i] = = Pattern[j]? j + 1:0;}} int main () {int cnt = 1;while (scanf ("%d", &m)!=eof,m) {scanf ("%s", pattern), Get_next ();p rintf ("Test case #%d\n", cnt+ +);/** * Iterates over the next array. * Output Cyclic section number >=2 */int i;for (i = 0; I <= m; ++i) {if (nnext[i] = = 0) {continue;} int len = i-nnext[i];if (I%len = = 0) {printf ("%d%d\n", I,i/len);}} printf ("\ n");}}

(KMP 1.5) HDU 1358 Period (use next array to find the smallest loop-the number of cyclic sections for the I-character)