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

 #include<stdio.h>
 #include<iostream>
 #include<string.h>
 #include<algorithm>
 using namespace std;

 const int MAXN=;
 char T[MAXN];
 int next[MAXN];
 int n;
 void getNext()
 {
     int j,k;
     j=;
     k=-;
     next[]=-;
     while(j < n)
     {
         if(k==-||T[j]==T[k])
         {
             j++;
             k++;
             if(j%(j-k)==&&j/(j-k)>)
               printf("%d %d\n",j,j/(j-k));
             next[j]=k;
         }
         else k=next[k];
     }
 }
 int main()
 {

     int iCase=;
     while(scanf("%d",&n),n)
     {
         iCase++;
         scanf("%s",&T);
         printf("Test case #%d\n",iCase);
         getNext();
         printf("\n");
     }
     return ;
 }