String（hdu 4821 字符串Hash+Map）

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5324    Accepted Submission(s): 1582

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

2

 

Source

2013 Asia Regional Changchun 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
#define ull unsigned long long   //溢出部分自动取余
using namespace std;

int m,l;
int base=133;
const int maxn=1e5+5;
ull p[maxn],Hash[maxn];    //用ull类型
map<ull,int> ma;
string name;

void init(){
    p[0]=1;
    for(int i=1;i<=maxn-1;i++) p[i]=p[i-1]*base;
}    //初始化

ull get(int left,int right,ull g[]){
    return g[right]-g[left-1]*p[right-left+1];
}

int main(){
    init();
    while(cin>>m>>l){
        cin>>name;
        int len=name.size();
        name=" "+name;
        Hash[0]=0;
        for(int i=1;i<=len;i++){
            Hash[i]=Hash[i-1]*base+name[i]-'a';   //对字符串进行Hash
        }
        int res=0;
        for(int i=1;i<=l&&i+m*l<=len;i++){
            ma.clear();
            for(int j=i;j<i+m*l;j+=l){
                ull x=get(j,j+l-1,Hash);
                ma[x]++;
            }
            if(ma.size()==m) res++;
            for(int j=i+m*l;j+l-1<=len;j+=l){   //头部删除,尾部加入
                ull x=get(j,j+l-1,Hash);
                ma[x]++;
                ull y=get(j-m*l,j-m*l+l-1,Hash);
                ma[y]--;
                if(ma[y]==0) ma.erase(y);   //不要忘记删除
                if(ma.size()==m) res++;
            }
        }
        cout << res << endl;
    }
    return 0;
}