HDU 1358Period(KMP周期串)

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1993    Accepted Submission(s): 978

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

3aaa12aabaabaabaab0

 

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

 

   

             题目大意：给你一个串，如果它的前缀是周期大于1的串，便输出这个子串（输出i+周期）.

        

        解题思路：需要用到一个结论，一个串的最小循环节是(i-next[i])，如果i%(i-next[i])==0，那么说明是一个周期串。方法可以参考：KMP最小循环节讲解 最下面。

        

        题目地址：Period

AC代码：

#include<iostream>#include<cstring>#include<string>#include<cstdio>using namespace std;char a[1000005];int next[1000005],len;void getnext(){     int i,j;     next[0]=0,next[1]=0;     for(i=1;i<len;i++)     {          j=next[i];          while(j&&a[i]!=a[j])               j=next[j];          if(a[i]==a[j])               next[i+1]=j+1;            else               next[i+1]=0;     }}int main(){     int cas=0;     while(scanf("%d",&len))     {          if(len==0) break;          scanf("%s",a);          getnext();          printf("Test case #%d\n",++cas);          for(int i=2;i<=len;i++)               if(i%(i-next[i])==0)                 if(i/(i-next[i])>1)                  printf("%d %d\n",i,i/(i-next[i]));          puts("");     }     return 0;}