String

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=95149#problem/I

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

HINT

题意

给你一个字符串，让你找到有多少个长度为m*l的子串，由m个长度为l的不同的串构成的

题解：

hash一下之后，就直接暴力找就好了

暴力得用类似滑块一样优化一下就好了

代码：

#include<stdio.h>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
#define maxn 100005
ll h[maxn*4];
ll h2[maxn*4];
char str[maxn*4];
int n,len,k,s1,s3,vis[maxn],sum[maxn];
ll N=1000000007;
ll p=31;
ll powp[maxn*4];

void get_hash()
{
    h[0]=(ll)str[0];
    for(int i=1;i<n;i++)
    {
        h[i]=(h[i-1]*p+(ll)str[i]);
        while(h[i]<0)
            h[i]+=N;
        if(h[i]>=N)
            h[i]%=N;
    }
    powp[0]=1LL;
    for(int i=1;i<n;i++)
    {
        powp[i]=powp[i-1]*p;
        while(powp[i]<0)
            powp[i]+=N;
        if(powp[i]>=N)
            powp[i]%=N;
    }
}
ll gethash(int l,int r)
{
    if(!l)
        return h[r];
    ll ans=h[r]-h[l-1]*powp[r-l+1];
    if(ans<0)
        ans%=N;
    if(ans<0)
        ans+=N;
    if(ans>=N)
        ans%=N;
    return ans;
}


map<ll ,int> H;
int main()
{
    int M,L;
    while(scanf("%d%d",&M,&L)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        memset(sum,0,sizeof(sum));
        scanf("%s",str);
        len = strlen(str);
        n = len;
        get_hash();
        /*
        int ans = len-M*L+1;
        for(int i=0;i<L;i++)
        {
            H.clear();
            int flag=0;
            for(int j=0;i+(j+1)*L<=len;j++)
            {
                ll pp=gethash(i+j*L,i+(j+1)*L-1);
                if(H[pp])
                {
                    vis[i+(H[pp]-1)*L]=1;
                    vis[i+j*L]=1;
                }
                H[pp]=j+1;
            }
            if(vis[i])sum[i]=1;else sum[i]=0;
            for(int j=1;i+j*L+L<=len;j++)
            {
                sum[i+j*L]=sum[i+(j-1)*L];
                if(vis[i+j*L])
                {
                    sum[i+j*L]++;

            }
            if(i+M*L<=len&&sum[i+(M-1)*L]!=0) ans--;
            for(int j=M;i+j*L+L<=len;j++)
                if(sum[i+(j-M)*L]!=sum[i+j*L]) ans--;
        }
        */
        int ans = 0;
        for(int i=0;i<L;i++)
        {
            H.clear();
            for(int j=0;j<M&&i+(j+1)*L-1<len;j++)
            {
                //cout<<i+j*L<<" "<<i+(j+1)*L-1<<" 1"<<" "<<gethash(i+j*L,i+(j+1)*L-1)<<endl;
                H[gethash(i+j*L,i+(j+1)*L-1)]++;
            }
           //cout<<H.size()<<endl;
            if(H.size()==M)ans++;
            for(int j=M;i+(j+1)*L-1<len;j++)
            {
                //cout<<i+j*L<<" "<<i+(j+1)*L-1<<" 2"<<" "<<gethash(i+(j-M)*L,i+(j+1-M)*L-1)<<endl;
                H[gethash(i+(j-M)*L,i+(j+1-M)*L-1)]--;
                if(H[gethash(i+(j-M)*L,i+(j+1-M)*L-1)]==0)
                    H.erase(gethash(i+(j-M)*L,i+(j+1-M)*L-1));
                H[gethash(i+j*L,i+(j+1)*L-1)]++;
                //cout<<H.size()<<endl;
                if(H.size()==M)ans++;
            }
        }
        printf("%d\n",ans);
    }
}