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HDOJ 4821 String

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HDOJ 4821 String

字符串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 Description Given 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 "a".

Your task is to calculate the number of different “recoverable” substrings of S.
Input The 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.
Output For each test case, output the answer in a single line.
Sample Input

3 3
abcabcbcaabc

Sample Output

Source 2013 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;
}