(KMP 1.5)hdu 1358 Period(使用next數組來求最小迴圈節——求到第i個字元的迴圈節數)，hdu1358

(KMP 1.5)hdu 1358 Period(使用next數組來求最小迴圈節——求到第i個字元的迴圈節數)，hdu1358

題目：

Period Time 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 has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK , that 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 two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.
 
OutputFor each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
 
Sample Input

3aaa12aabaabaabaab0

 
Sample Output

Test case #12 23 3Test case #22 26 29 312 4

 
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題目分析：

         KMP。簡單題。這一道題其實和KMP 1.4那道題的思想是一樣的。

代碼如下：

/* * hdu1358.cpp * *  Created on: 2015年4月18日 *      Author: Administrator */#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1000001;int m;//目標串的長度char pattern[maxn];//模式串int nnext[maxn];//next數組.直接起next可能會跟系統中預定的重名./*O(m)的時間求next數組*/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();printf("Test case #%d\n",cnt++);/** * 遍曆next數組。 * 輸出迴圈節數>=2的情況 */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");}}