hash-白兔的字符串

Posted 2020-12-17 jxust-biao

^{https://ac.nowcoder.com/acm/problem/15253}
太简单了，直接上代码吧！

#include <iostream>
#include <cstring>
#include <cmath>
#include <stdio.h>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <iomanip>
#define rep(i,a,b) for(int i = a; i <= b ; i ++ )
#define pre(i,a,b) for(int i = b ; i >= a ; i --)
#define ll long long
#define inf 0x3f3f3f3f
#define ull unsigned long long
#define ios ios::sync_with_stdio(false),cin.tie(0)
using namespace std;
typedef pair<int,int> PII ;
const int N = 2e6 + 10 ;
ull base[N];
ull hs[N];
char T[N],s[N];
vector<ull> v;
ull n;

ull gethash(int i,int j){//get hash of [i,j]
    return hs[j]-hs[i-1]*base[j-i+1];
}

bool judge(int i,int j){
    ull x=gethash(i,j);//the part of son‘s hash
    int id=lower_bound(v.begin(),v.end(),x)-v.begin();//found of two found
    if(id==v.size())return false;//if not found
    else return v[id]==x; //judge whether be found
}

int main()
{
    ios;
    base[0] =1 ;
    for(int i = 1 ; i < N ; i ++)
        base[i]= base[i-1]*131;
    cin>>(T+1);
    ull lent=strlen(T+1);
    for(int i=1;i<=lent;i++) T[i+lent]=T[i];// one plus
    for(int i=1;i<=2*lent;i++) hs[i]=hs[i-1]*131+(T[i]-‘a‘+1);
    for(int i=1;i<=lent;i++) v.push_back(gethash(i,i+lent-1));
    sort(v.begin(),v.end());//sort the result of hash

    cin>>n;
    while(n--)
    {
        cin>>(s+1);
        ull lens=strlen(s+1);
        int cnt=0;
        for(int i=1;i<=lens;i++){
            hs[i]=hs[i-1]*131+(s[i]-‘a‘+1);//do hash
        }
        for(int i=1;i+lent-1<=lens;i++){
            if(judge(i,i+lent-1)){
                    cnt++;
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}