hdu 1358 Period (KMP)

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2402    Accepted Submission(s): 1191

Problem Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

 

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

 

Output

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

 

Sample Input

3 aaa 12 aabaabaabaab 0

 

Sample Output

Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4

 

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//109MS    5116K    636 B    G++
/*

    题意：
        给出一字符串，找出由两个或以上相同的子字符串组成的前缀，
    输出前缀长度及其中相同的子字符串数
    
    KMP：
       这是一题对next数组的应用的题，首先常规求出next数组，然后 
            
          式子 ：  i%temp==0 && i/temp>1
       
       为关键的判断句，具体理解可以找实例试试 
        
*/
#include<stdio.h>
int next[1000005];
int Get_next(char c[],int n)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<n){
        if(c[i]==c[j] || j==-1){
            i++;j++;next[i]=j;
        }
        else j=next[j];
    }
}
int main(void)
{
    int n,k=1;
    char c[1000005];
    while(scanf("%d",&n),n)
    {
        scanf("%s",c);
        Get_next(c,n);
        printf("Test case #%d
",k++);
        for(int i=1;i<=n;i++)
        {
            int temp=i-next[i];
            if(i%temp==0 && i/temp>1)
                printf("%d %d
",i,i/temp);
        }
        printf("
");
    }
    return 0;
}