HDOJ 4821 String

HDOJ 4821 String

String hash

String Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 697 Accepted Submission (s): 190


Problem DescriptionGiven a string S and two integers L and M, we consider a substring of S as "recoverable" if and only if
(I) It is of length M * L;
(Ii) It can be constructed by concatenating M "diversified" substrings of S, where each of these substrings has length L; two strings are considered as "diversified" if they don't have the same character for every position.

Two substrings of S are considered as "different" if they are cut from different part of S. for example, string "aa" has 3 different substrings "aa", "a" and "".

Your task is to calculate the number of different "recoverable" substrings of S.
InputThe input contains multiple test cases, proceeding to the End of File.

The first line of each test case has two space-separated integers M and L.

The second ine of each test case has a string S, which consists of only lowercase letters.

The length of S is not larger than 10 ^ 5, and 1 ≤ M * L ≤ the length of S.
OutputFor each test case, output the answer in a single line.
Sample Input

3 3abcabcbcaabc

Sample Output

2

Source2013 Asia Regional Changchun

#include 
 
  #include 
  
   #include 
   
    #include #include 
    using namespace std;typedef unsigned long long int ull;const int maxn=100100;int L,M;char str[maxn];ull xp[maxn],hash[maxn];map
     
       ck;void init(){xp[0]=1;for(int i=1;i
      
       =0;i--) { hash[i]=hash[i+1]*175+(str[i]-'a'+1); } int ans=0; for(int i=0;i
       
         i+(j+1)*L-1 duan++; ull hahashsh=get_hash(i+j*L,L); ck[hahashsh]++; if(duan>=M) { if(duan>M) { /// M+1 ago : i+(j+1)*L-L*(M+1) ull Mago=get_hash(i+(j+1)*L-L*(M+1),L); if(ck[Mago]) { ck[Mago]--; if(ck[Mago]==0) ck.erase(Mago); } } if(ck.size()==M) ans++; } } } printf("%d\n",ans); } return 0;}