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=1000010;
char T[MAXN];
int next[MAXN];
int n;
void getNext()
{
    int j,k;
    j=0;
    k=-1;
    next[0]=-1;
    while(j < n)
    {
        if(k==-1||T[j]==T[k])
        {
            j++;
            k++;
            if(j%(j-k)==0&&j/(j-k)>1)
              printf("%d %d
",j,j/(j-k));
            next[j]=k;
        }
        else k=next[k];
    }
}
int main()
{
   
    int iCase=0;
    while(scanf("%d",&n),n)
    {
        iCase++;
        scanf("%s",&T);
        printf("Test case #%d
",iCase);
        getNext();
        printf("
");
    }
    return 0;
}