2013
11-10

# Period

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

The input 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.

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.

3
aaa
12
aabaabaabaab
0

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

//* @author:
import java.util.Scanner;

public class Main {
static int next[]=new int[1000010];

public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int n;
String input = "";
int j=0;
int k=0;
while(scn.hasNext()){
n=scn.nextInt();
input = scn.next();
if(n==0){
break;
}
getNext(input,n);
System.out.printf("Test case #%d\n",++k);

/* 挨个试验 */
for (int i = 2; i <= n; i++){
/* 计算首尾重复子串的长度 */
j = i - next[i];
/* 串满足重复性质且重复子串不为本身 */
if (i % j == 0 && i / j > 1){
System.out.printf("%d %d\n", i, i / j);
}
}
System.out.printf("\n");
}

}

public static void getNext(String T,int len) {//建立模式串T的next[]表
int i = 0;
int j = next[0] = -1;

while (i< len)
if (0 > j || T.charAt(i) == T.charAt(j)) {//匹配
j++; i++;
next[i] = j;
}else//失配
j = next[j];
}
}

1. 有限自动机在ACM中是必须掌握的算法，实际上在面试当中几乎不可能让你单独的去实现这个算法，如果有题目要用到有限自动机来降低时间复杂度，那么这种面试题应该属于很难的级别了。

2. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n