POJ 1961 Period( KMP )*

Period

Time Limit: 3000MS
Memory Limit: 30000K

Total Submissions: 12089
Accepted: 5656

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

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::这道题跟poj2406差不多。题意：对于选定长度i（前i个字符,2<=i<=N），求出对应循环节k>1的情况

比如： aabaabaabaab

当i=2时, a ，a循环节有2个

当i=6时, aab,aab循环节有2个

当i=9时, aab,aab,aab循环节有3个

当i=12时，aab,aab,aab,aab循环节有4个

假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]

证明见POJ 2406 Power Strings (KMP)

代码：

   1: #include <iostream>

   2: #include <cstdio>

   3: #include <cstring>

   4: #include <algorithm>

   5: using namespace std;

   6: const int maxn=1e6;

   7: char s[maxn+10];

   8: int next[maxn+10];

   9:  

  10: void get_next(char s[],int len)

  11: {

  12:     int i=0,j=-1;

  13:     next[0]=-1;

  14:     while(i<len)

  15:     {

  16:         if(j==-1||s[i]==s[j]) {i++; j++; next[i]=j;}

  17:         else j=next[j];

  18:     }

  19: }

  20:  

  21: int main()

  22: {

  23:     int n,cas=1;

  24:     while(scanf("%d",&n)>0&&n)

  25:     {

  26:         scanf("%s",s);

  27:         printf("Test case #%d\n",cas++);

  28:         get_next(s,n);

  29:         for(int i=2; i<=n; i++)//枚举长度i

  30:         {

  31:             if(i%(i-next[i])==0&&i!=(i-next[i]))

  32:                 printf("%d %d\n",i,i/(i-next[i]));

  33:         }

  34:         printf("\n");

  35:     }

  36:     return 0;

  37: }