HDU 1358 period (KMP cycle string)

Period

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 1993 accepted submission (s): 978


Problem descriptionfor each prefix of a given string s with n characters (each character has an ascii code between 97 and 126, aggressive), 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 AK, that is a concatenated K times, for some string. of course, we also want to know the period K.

Inputthe 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.

Outputfor 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

Give you a string. If its prefix is a string with a period greater than 1, output this substring (output I + period ). solution: we need to use a conclusion. The smallest cyclic section of a string is (I-next [I]). If I % (I-next [I]) = 0, it indicates a periodic string. For more information, see the KMP least cycle section. Subject address: Period
AC code:

#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;}