hdu1358 Period

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2524    Accepted Submission(s): 1245

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
  
  
  
  
  

  

  
   
   
   
   

 

一直在思索这个题目，发现自己对next数组理解还是不深刻。

举个简单的例子：

i	0	1	2	3	4	5	6	7	8	9	10	11	12	13
str	a	a	b	a	a	b	a	a	b	a	a	b	\0
next[i]	-1	0	1	0	1	2	3	4	5	6	7	8	9

GetNext函数的时候可以取得12的next值。i为11时，比较不相等，返回到8，意味着在11的时候，比较不相等，跳转至8的位置。为什么跳转到8呢？因为之前的都相等，而8和11位置处的关系不确定，所以跳至8的位置。当然，可以优化GetNext算法。

//============================================================================
// Name        : KMP_hdu1358.cpp
// Author      : vit
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

int i, j, m ,n;
char str[1000001];
int next[1000001];
int ch;
int no;

void GetNext() {
	j = 0;
	i = -1;
	next[0] = -1;
	while (j < n) {
		if (i == -1 || str[i] == str[j]) {
			i++, j++;
			next[j] = i;
		} else {
			i = next[i];
		}
	}
}

int main() {
	no = 1;
	while(cin >> n){
		if(n == 0)
			break;
		scanf("%s", str);
		GetNext();
		printf("Test case #%d\n", no);
		no++;
		for(i = 2; i <= n; i++){
			ch = i - next[i];
			if(i % ch == 0){//可以想成（i - 1） - 0 + 1
				if(i / ch > 1){//重复次数大于1
					printf("%d %d\n",i,i / ch);
				}
			}
		}
		printf("\n");
	}
	return 0;
}

//优化的算法

void GetNextval(){//用此算法生成的nextval数组，不能用上面的算法进行计算。原因很简单：把相同的串重复跳转判断的部分放在了nextval生成里面，所以减少KMP的比较次数的同时，也造成了无法找到重复串的次数。
	j = 0;
	i = -1;
	nextval[0] = -1;
	while(j < n){
		if(i == -1 || str[i] == str[j]){
			i++, j++;
			if(str[i] == str[j])
				nextval[j] = i;
			else
				nextval[j] = nextval[i];
		} else
			i = nextval[i];
	}
}