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 DescriptionFor 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 AK , that is A concatenated K times, for some string A. 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

                 題目大意：給你一個串，如果它的首碼是周期大於1的串，便輸出這個子串（輸出i+周期）.                解題思路：需要用到一個結論，一個串的最小迴圈節是(i-next[i])，如果i%(i-next[i])==0，那麼說明是一個周期串。方法可以參考：KMP最小迴圈節講解 最下面。                題目地址：Period
AC代碼：

#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;}