poj 1961 Period

Posted 2020-10-27 adelalove

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：（注意输出）
给定字符串S，求1-i的字符串的循环节（循环节长度要大于1）
解：
假设要求整个字符串的循环节 poj 2406

len为整个字符串的长度

  想一想next[i]的含义

  每次匹配失败后，都要回到next[i]开始重新匹配

  也就是说  next[i]+1 -> len  这段字符 (令k为这段字符的长度)和

  1 -> 1+k  这段字符 是相同的

  OK，如果(len%(len-next[len])==0)  ==>  存在 len/(len-next[len]) 段 解

  否则只有1段

那么求1-i的字符串的循环节同理

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<string>
 7 #include<queue>
 8 #include<map>
 9 using namespace std;
10 const int N=1e7;
11 char s[N];
12 int len,ne[N],j,k;
13 int main()
14 {
15     while(scanf("%d",&len)!=EOF)
16     {
17         if(!len) break;
18         scanf("%s",s+1);k++;
19         printf("Test case #%d\n",k);
20         for(int i=0;i<=len;++i) ne[i]=0;
21         for(int i=2;i<=len;++i)
22         {
23             j=ne[i-1];
24             while(j && s[i]!=s[j+1]) j=ne[j];
25             if(s[i]==s[j+1]) ne[i]=j+1;
26         }
27         for(int i=1;i<=len;++i)
28         if(i%(i-ne[i])==0 && i/(i-ne[i])>1) 
29          printf("%d %d\n",i,i/(i-ne[i]));
30         printf("\n");
31     }
32     return 0;
33 }

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