KMP (longest prefix and suffix)

http://acm.hdu.edu.cn/showproblem.php?pid=1358

Period

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

 

Recommend

JGShining

 

The meaning of problems: a string, all the output Q: How much of the length (total length or less), can be from a minimum of two identical sub-strings

 Note: The title alone is seeking the longest prefix and suffix, it is best not to carry out next array optimization (matching problem can be used)

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
char a[1000009];


void getnext(char *a , int len , int *next)
{
    next[0] = -1 ;
    int k = -1 , j = 0;
    while(j < len)
    {
        if(k == -1 || a[j] == a[k])
        {
            k++;
            j++;
            next[j] = k ;
        }
        else
        {
            k = next[k];
        }
    }
}

int main()
{
    intn-, ANS = 0 ;
     the while (~ Scanf ( " % D " , & n-) && n-) 
    { 
        int Next [ 1000009 ]; 
        Scanf ( " % S " , A); 
        the printf ( " the Test Case #% D \ n- " , ++ ANS); 

        Memset (next, 0 , the sizeof (next)); 
        GetNext (A, n-, next); // find next array 
        for ( int I = 2 ; I <= n-; I ++) // iterate through each 2 is greater than the string 
        {
             //i% i (i - next [ i]) indicates that the string consists of a number (greater than one) the same strings 
            IF (Next [I]> = . 1 && I% (I - Next [I]) = = 0 ) 
            { 
                int L = I - Next [I]; 
                the printf ( " % D% D \ n- " , I, I / L); // I / L find the same number of a character string composed of the same 
            } 
        } 
        the printf ( " \ n- " ); 
    } 

    return  0 ; 
}