codeforces 55d//Beautiful numbers// Codeforces Beta Round #51

Posted 2021-01-09 gaudar

题意：一个数能整除它所有的位上的数字（除了0），统计这样数的个数。

注意离散化，为了速度更快需存入数组查找。

不要每次memset，记录下已有的长度下符合条件的个数。

数位dp肯定是从高位到低位。

记录数字已经有多大，还有lcm，递归传下去。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
using namespace std;
const double EPS=1e-8;
const int SZ=5050,INF=0x7FFFFFFF;
typedef long long lon;
lon dp[20][3000][50],index[3000];//2520
vector<lon> ls;

lon gcd(lon x,lon y)
{
    if(x<y)swap(x,y);
    for(;;)
    {
        lon rem=x%y;
        if(rem==0)return y;
        x=y;
        y=rem;
    }
}

lon lcm(lon x,lon y)
{
    return x*y/gcd(x,y);
}

void getid()
{
    for(lon i=0;i<ls.size();++i)
    {
        index[ls[i]]=i;
    }
}

void lsh()
{
    for(lon i=1;i<pow(2,10);++i)
    {
        lon cur=1;
        for(lon j=0;j<9;++j)
        {
            if(i&(1<<j))cur=lcm(cur,j+1);
        }
        ls.push_back(cur);
    }
    sort(ls.begin(),ls.end());
    ls.erase(unique(ls.begin(),ls.end()),ls.end());
    getid();
}

lon dfs(lon pos,lon cur,lon prelcm,lon limit,string &str)
{
    if(pos==str.size())return cur%prelcm==0;
    if(!limit&&dp[str.size()-pos-1][cur][index[prelcm]]!=-1)return dp[str.size()-pos-1][cur][index[prelcm]];
    lon res=0;
    lon up=limit?str[pos]-‘0‘:9;
    for(lon i=0;i<=up;++i)
    {
        lon curlcm=prelcm;
        if(i)curlcm=lcm(i,curlcm);
        res+=dfs(pos+1,(cur*10+i)%2520,curlcm,limit&&str[pos]-‘0‘==i,str);
    }
    if(!limit)dp[str.size()-pos-1][cur][index[prelcm]]=res;
    return res;
}

lon work(string &str)
{
    return dfs(0,0,1,1,str);
}

bool chk(string &str)
{
    lon num=0,curlcm=1;
    for(lon i=0;i<str.size();++i)
    {
        num=(num*10+str[i]-‘0‘)%2520;
        if(str[i]!=‘0‘)curlcm=lcm(curlcm,str[i]-‘0‘);
    }
    return num%curlcm==0;
}

int main()
{
    std::ios::sync_with_stdio(0);
    //freopen("d:\1.txt","r",stdin);
    lon casenum;
    cin>>casenum;
    lsh();
    memset(dp,-1,sizeof(dp));
    for(lon time=1;time<=casenum;++time)
    {
        string ll,rr;
        cin>>ll>>rr;
        lon res=work(rr)-work(ll);
        if(chk(ll))++res;
        cout<<res<<endl;
    }
    return 0;
}