poj--1961--Period

亦凉 2022-06-09 01:11 159阅读 0赞

Period

<table> 
 <tbody> 
 <tr> 
 <td>Time Limit:&nbsp;3000MS</td> 
 <td>&nbsp;</td> 
 <td>Memory Limit:&nbsp;30000K</td> 
 </tr> 
 <tr> 
 <td>Total Submissions:&nbsp;18603</td> 
 <td>&nbsp;</td> 
 <td>Accepted:&nbsp;9044</td> 
 </tr> 
 </tbody> 
</table>

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 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

题意：

求字符串的每个前缀的大于等于2的循环节长度

思路：

考察KMP算法中对next数组的理解，如果对KMP算法比较了解的话，很容易就能写出来

代码：

C++ Code

<table style="width:100%; font-family:'Consolas','Courier New'; font-size:12px; vertical-align:text-top; line-height:15px"> 
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 <td style="background-color:rgb(239,239,239); width:100%; vertical-align:text-top"> 
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 #include&lt;cstdio&gt; 
 
 using&nbsp; 
 namespace&nbsp;std; 
 
 
 const&nbsp; 
 int&nbsp;MAXN&nbsp;=&nbsp; 
 1000010; 
 
 char&nbsp;str[MAXN]; 
 
 int&nbsp;next[MAXN]; 
 
 int&nbsp;n; 
 
 void&nbsp;getNext() 
 { 
 &nbsp;&nbsp;&nbsp;&nbsp; 
 int&nbsp;i,&nbsp;j; 
 &nbsp;&nbsp;&nbsp;&nbsp;i&nbsp;=&nbsp; 
 0; 
 &nbsp;&nbsp;&nbsp;&nbsp;j&nbsp;=&nbsp;- 
 1; 
 &nbsp;&nbsp;&nbsp;&nbsp;next[ 
 0]&nbsp;=&nbsp;- 
 1; 
 &nbsp;&nbsp;&nbsp;&nbsp; 
 while(str[i]&nbsp;!=&nbsp; 
 '\0') 
 &nbsp;&nbsp;&nbsp;&nbsp;{ 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
 if(j&nbsp;==&nbsp;- 
 1&nbsp;||&nbsp;str[i]&nbsp;==&nbsp;str[j]) 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{ 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;i++; 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;j++; 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
 if(i&nbsp;%&nbsp;(i&nbsp;-&nbsp;j)&nbsp;==&nbsp; 
 0&nbsp;&amp;&amp;&nbsp;i&nbsp;/&nbsp;(i&nbsp;-&nbsp;j)&nbsp;&gt;&nbsp; 
 1) 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;printf( 
 "%d&nbsp;%d\n",&nbsp;i,&nbsp;i&nbsp;/&nbsp;(i&nbsp;-&nbsp;j)); 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;next[i]&nbsp;=&nbsp;j; 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;} 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
 else&nbsp;j&nbsp;=&nbsp;next[j]; 
 &nbsp;&nbsp;&nbsp;&nbsp;} 
 } 
 
 int&nbsp;main() 
 { 
 &nbsp;&nbsp;&nbsp;&nbsp; 
 int&nbsp;iCase&nbsp;=&nbsp; 
 0; 
 &nbsp;&nbsp;&nbsp;&nbsp; 
 while(scanf( 
 "%d",&nbsp;&amp;n),&nbsp;n) 
 &nbsp;&nbsp;&nbsp;&nbsp;{ 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;iCase++; 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;scanf( 
 "%s",&nbsp;&amp;str); 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;printf( 
 "Test&nbsp;case&nbsp;#%d\n",&nbsp;iCase); 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;getNext(); 
 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;printf( 
 "\n"); 
 &nbsp;&nbsp;&nbsp;&nbsp;} 
 &nbsp;&nbsp;&nbsp;&nbsp; 
 return&nbsp; 
 0; 
 } 
 </div> </td> 
 </tr> 
 </tbody> 
</table>