poj 1961 (求字符串中的重复子串)

Sample Input

3
aaa
12
aabaabaabaab
0
Sample Output

Test case #1
2 2
3 3

Test case #2
2 2 //aa有2个a
6 2 //aabaab有2个aab
9 3
12 4

0  1  2 3 4 5 6 7 8 9 10 11

a  a b a a b  a a b a a b     的next数组为
-1 0 1 0 1 2 3 4 5 6 7 8 9

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<algorithm>
 5 using namespace std;
 6
 7 const int MAXN=1000010;
 8 char T[MAXN];
 9 int next[MAXN];
10 int n;
11 void getNext()
12 {
13     int j,k;
14     j=0;
15     k=-1;
16     next[0]=-1;
17     while(j < n)
18     {
19         if(k==-1||T[j]==T[k])
20         {
21             j++;
22             k++;
23             if(j%(j-k)==0&&j/(j-k)>1)
24               printf("%d %d\n",j,j/(j-k));
25             next[j]=k;
26         }
27         else k=next[k];
28     }
29 }
30 int main()
31 {
32
33     int iCase=0;
34     while(scanf("%d",&n),n)
35     {
36         iCase++;
37         scanf("%s",&T);
38         printf("Test case #%d\n",iCase);
39         getNext();
40         printf("\n");
41     }
42     return 0;
43 }