hdu 4821 ||2013年长春站J题 字符串哈希+map的应用

http://acm.hdu.edu.cn/showproblem.php?pid=4821

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 3abcabcbcaabc

 

Sample Output

2

题目大意：

                  给定一个字符串在其中找出子串长度 m*l，并且该字串可分为m段每段长度均为l，求有多少个这样的子串。

解题思路：

                 利用哈希算出每个子串的哈希值，利用map，注意不能直接暴力一遍，会超时，技巧在代码。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <map>//#define debugusing namespace std;typedef unsigned long long ULL;const int SIZE = 100003;const int SEED = 13331;const int MAX_N = 110000 + 10;char s[MAX_N];struct HASH{    ULL H[MAX_N];    ULL XL[MAX_N];    int len;    HASH(){}    void build(char *s){        len=strlen(s);        H[len]=0;        XL[0]=1;        for (int i=len-1;i>=0;i--){            H[i]=H[i+1]*SEED+s[i];            XL[len-i]=XL[len-i-1]*SEED;        }    }    ULL hash(int i,int L){        return H[i]-H[i+L]*XL[L];    }}hs;char a[100005];int n,m,cnt;map<ULL,int> p;int main(){    while(~scanf("%d%d",&n,&m))    {        scanf("%s",a);        hs.build(a);        int len=strlen(a);        cnt=0;        for(int i=0;i<m&&i<=(len-n*m);i++)        {            p.clear();            for(int j=i;j<i+n*m;j+=m)                p[hs.hash(j,m)]++;            if(p.size()==n)                cnt++;            for(int j=i+n*m;j<=len-m;j+=m)            {                p[hs.hash(j-n*m,m)]--;                if(p[hs.hash(j-m*n,m)]==0)                    p.erase(hs.hash(j-m*n,m));                p[hs.hash(j,m)]++;                if(p.size()==n)                    cnt++;            }        }        printf("%d\n",cnt);    }    return 0;}

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 3abcabcbcaabc

 

Sample Output

2