HDU 4821 String 字符串hash

String

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

 

题意：

　　给你M和L，和一个字符串S。

　　要求找出S的子串中长度为L*M，并且可以分成M段，每段长L，并且M段都不相同的子串个数。

题解：

　　枚举起点

　　hash每个前缀串

　　那么一段子串的hash值就可以快速求出

　　twopointsO(n)求出长度M*L，,M个连续串是否相同

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 5e5+10, MM = 1e3+20,inf = 2e9;

const LL mod = 10000019ULL;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

map<ULL,int  > s;
ULL bhas[N],has[N],sqr[N];
int M,L;
char sa[N];
int vis[N];
int main() {
    sqr[0] = 1;
    for(int i = 1; i < N; ++i) sqr[i] = sqr[i-1] * mod;
    while(scanf("%d%d",&M,&L)!=EOF) {
        scanf("%s",sa+1);
        int n = strlen(sa+1);
        has[0] = 0;
        for(int i = 1; i <= n; ++i) {
            has[i] = has[i-1] * mod + sa[i] - 'a' + 1;
        }
        int ans = 0;
        for(int i = 1; i <= L && i + M * L - 1 <= n; ++i) {
            int cnt = 0;
            s.clear();
            int ll = 1,rr = 0;
            for(int j = i; j + L - 1 <= n; j += L) {
                int l = j, r = j + L - 1;
                ULL now = has[r] - has[l-1]*sqr[L];
                if(s[now] == 0){
                    bhas[++rr] = now;
                    if(rr - ll + 1 >= M)ans+=1;
                    s[now] = 1;
                    continue;
                }
                else {
                    while(ll <= rr && bhas[ll]!=now) {
                        s[bhas[ll++]] = 0;
                    }
                    s[bhas[ll++]] = 0;
                    bhas[++rr] = now;
                    if(rr - ll + 1 >= M)ans+=1;
                    s[now] = 1;
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}