Poj1961--Period（Kmp, Next数组求循环节长度 && 出现次数） & Poj 2406--Power Strings

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 14657		Accepted: 6967

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 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

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

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

Source

Southeastern Europe 2004

RE: 求长度为（0 ~ i）的循环节出现的次数; (注意循环节和（前、后缀）的区别)；

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 char a[1000100]; int lena, p[1000100];
 6 void Getp()
 7 {
 8     int i = 0, j = -1;
 9     p[i] = j;
10     while(i < lena)
11     {
12         if(j == -1 || a[i] == a[j])
13         {
14             i++; j++;
15             p[i] = j;
16         }
17         else
18             j = p[j];
19     }
20 }
21 int main()
22 {
23     int t, csse = 1;; 
24     while(~scanf("%d", &t), t)
25     {
26         scanf("%s", a);
27         lena = strlen(a); 
28         Getp();
29         printf("Test case #%d
", csse++);    
30         for(int i = 1; i <= lena; i++)
31         {
32             //int length = i - p[i];              //循环节长度；
33             if(i % (i - p[i]) == 0 && p[i] != 0)  //判断是否为循环节；
34                 printf("%d %d
", i, i/(i-p[i]));
35         }
36         printf("
");
37     }
38     return 0;
39 }

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37625		Accepted: 15562

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

RE: 求最大循环节出现次数；循环节判断 len%(len-p[len]) == 0 && p[len] != 0; 最大循环节一般出现在 p[len] 处；

Poj1961--Period（Kmp, Next数组求循环节长度 && 出现次数） & Poj 2406--Power Strings

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