poj------2406 Power Strings
A - Power Strings 难度：☆☆
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2406
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
3Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
from http://poj.org/problem?id=2406
考察kmp算法中的next数组....
next的两种表示方法，第一种是前缀next[]数组...
while(i<len)
{
if(-1==j||ss[i]==ss[j])
{
i++;
j++;
next[i]=j;
}
else j=next[j];
}另种表示方法文为：
while(i<len)
{
if(j==-1||ss[i]==s[j])
{
i++:
j++;
if(ss[i]==ss[j])
{
next[i]=next[j];
}
else next[i]=j;
}
else j=next[j];
}这道题是考察有多少个重复的最大n所以我们不妨看他的回溯长度len_D=len(已经匹配的位置) --next[len](对应部分的匹配值);
Java代码：
1 //package dek0;
2
3 import java.util.Scanner;
4
5 public class Main {
6
7 public static void main(String args[])
8 {
9 Scanner reader = new Scanner(System.in);
10 String ss="";
11 while(reader.hasNext())
12 {
13 ss=reader.next();
14 if(ss.charAt(0)=='.') break;
15 int len=ss.length();
16 int next[]=new int [len+1];
17 next[0]=-1;
18 int i=0,j=-1;
19 while(i<len)
20 {
21 if(j==-1||ss.charAt(i)==ss.charAt(j))
22 {
23 i++;
24 j++;
25 /* if(ss.charAt(i)==ss.charAt(j))
26 next[i]=next[j];
27 else next[i]=j;*/
28 next[i]=j;
29 }
30 else j=next[j];
31 }
32 if(len%(len-next[len])==0)
33 System.out.println(len/(len-next[len]));
34 else
35 System.out.println(1);
36 }
37 }
38 }