HDU1358 字符串求周期 KMP算法

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本文介绍了一种利用KMP算法来寻找字符串中所有前缀的最大周期及其对应的重复次数的方法。通过对KMP失败函数的计算与扩展应用，该算法能够高效地解决这一问题。

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209 Accepted Submission(s): 92

Problem Description

For 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 A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The 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.

Output

For 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

aaa

aabaabaabaab

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

求字符串中存在周期的子串的最大循环次数 ,KMP 算法的应用之一 .

代码如下 :

#include<stdio.h> #include<string.h> int fail[1000005]; char c[1000005]; int period[1000005]; int n; void comfail(char p[]) { int i,j; j=0; for (i=1; i<=n; ++i) { fail[i]=j; while (j>0&&p[i]!=p[j]) j=fail[j]; j++; } } int main() { int i,j; int cas=1,deal; while (scanf("%d",&n)&&n) { memset(c,0,sizeof(c)); memset(period,0,sizeof(period)); getchar(); for (i=1; i<=n; ++i) scanf("%c",&c[i]); comfail(c); fail[1]=1; for (i=2; i<n; ++i) { if (fail[i]<fail[i+1]) { deal=i-1; for (j=i; j<=n; ++j) { if (j%deal==0&&c[j]==c[deal]) period[j]=j/deal; if (fail[j]>fail[j+1]) break; } i=j; } } printf("Test case #%d/n",cas++); for (i=2; i<=n; ++i) { int t=i-fail[i]; if (i%t==0&&i!=t&&c[t]==c[i]) printf("%d %d/n",i,i/t); } printf("/n"); } return 0; }