poj 2406:Power Strings（KMP算法，next[]数组的理解）
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 30069 | Accepted: 12553 |
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
KMP算法,next[]数组的理解。
这道题考察的是对next[]数组的理解。如果 i 能整除 (i-next[i]) 的话，证明在这个字符之前，是一个由同一前缀重复连接构成的字符串。例如：aabaabaabc，c位置的下标 i 就可以整除他的下标和next数组值的差(i-next[i])，则它之前的字符串aabaabaab就是一个重复字符串，构成它的前缀的长度就是 (i - next[i])。
注意：输入以一个'.'结束；注意这一组测试数据“aabaabaa”，WA的可以测试一下，正确结果应该为1。
如果觉得文字太枯燥可以猛戳这里：hdu 1358:Period（KMP算法，next[]数组的使用）
代码：
1 #include <stdio.h>
2
3 char s[1000001];
4 int next[1000001];
5 void GetNext(char a[],int next[],int n) //获得a数列的next数组
6 {
7 int i=0,k=-1;
8 next[0] = -1;
9 while(i<n){
10 if(k==-1){
11 next[i+1] = 0;
12 i++;k++;
13 }
14 else if(a[i]==a[k]){
15 next[i+1] = k+1;
16 i++;k++;
17 }
18 else
19 k = next[k];
20 }
21 }
22
23 int main()
24 {
25 while(scanf("%s",s)!=EOF){
26 if(s[0]=='.') break;
27 int i;
28 for(i=0;s[i];i++);
29 int len = i;
30 GetNext(s,next,len);
31 if(len%(len-next[len])==0)
32 printf("%d\n",len/(len-next[len]));
33 else
34 printf("1\n");
35 }
36 return 0;
37 }
Freecode : www.cnblogs.com/yym2013
