HDU 4821 String 字符串哈希

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 937    Accepted Submission(s): 270

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

 

Source

2013 Asia Regional Changchun

 

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(int i=0;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define DWN(i,h,l) for(int i=(h);i>=(l);--i)#define CLR(vis,pos) memset(vis,pos,sizeof(vis))#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LINF 1000000000000000000LLtypedef unsigned long long ull;const int maxn=1e5+100;ull base=31;ull nbase[maxn],nhash[maxn];char s[maxn];int M,L;map<ull,int> p;int main (){    nbase[0]=1;    FOR(i,1,maxn-1) nbase[i]=nbase[i-1]*base;    while(scanf("%d%d%s",&M,&L,s)!=EOF){        int len=strlen(s);        nhash[len]=0;        DWN(i,len-1,0)         nhash[i]=nhash[i+1]*base+s[i]-'a'+1;        int ans=0;        ull tmp;        for(int i=0;i<L && i+M*L<=len;i++){            p.clear();            for(int j=i;j<i+M*L;j+=L){                tmp=nhash[j]-nhash[j+L]*nbase[L];                p[tmp]++;            }            if(p.size()==M) ans++;            for(int j=i+M*L;j+L<=len;j+=L){                tmp=nhash[j-M*L]-nhash[j-(M-1)*L]*nbase[L];                p[tmp]--;                if(p[tmp]==0) p.erase(tmp);                tmp=nhash[j]-nhash[j+L]*nbase[L];                p[tmp]++;                if(p.size()==M) ans++;            }        }        cout<<ans<<endl;    }    return 0;}