Codeforces Beta Round #51 D. Beautiful numbers （数位dp）

D. Beautiful numbers

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 9 ·10¹⁸).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from l_i to r_i, inclusively).

Sample test(s)

input

1
1 9

output

9

input

1
12 15

output

2

题意：

求区间内满足能被每位非零数整除数的个数。数据范围：9*10^18

思路来源于：男神博客

思路：

lcm(1,2,3,4,5,6,7,8,9)=2^3*3^2*5*7=2520  公约数的总个数为4*3*2*2=48个

a%b=0 可推出 a%(k*b)%b=0。

所以可以先对2520取模，再对各位出现数字的lcm取模。

可以用记忆化搜索 dfs(pos,r,lcm,flag) 表示到当前位，前缀模2520为r，前缀的最小公倍数为lcm，上界标志，枚举这一位选什么来转移，递归算答案。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#include <sstream>
#define maxn 100005
#define MAXN 100005
#define mod 2520
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
using namespace std;

ll n,m,ans,tot,cnt;
ll dig[21],dp[21][2525][50],mp[2525],fac[21];

ll gcd(ll x,ll y)
{
    if(y==0) return x;
    return gcd(y,x%y);
}
ll LCM(ll x,ll y)
{
    if(y==0) return x;
    return x/gcd(x,y)*y;
}
ll dfs(ll pos,ll r,ll lcm,bool flag)
{
    if(pos==0)
    {
        if(r%lcm==0) return 1;
        return 0;
    }
    if(!flag&&dp[pos][r][mp[lcm]]!=-1) return dp[pos][r][mp[lcm]];
    ll i,j,t,best=0,ed,r1,lcm1;
    if(flag) ed=dig[pos];
    else ed=9;
    for(i=0;i<=ed;i++)
    {
        r1=(r+i*fac[pos-1])%mod;
        lcm1=LCM(lcm,i);
        t=dfs(pos-1,r1,lcm1,flag&&i==ed);
        best+=t;
    }
    if(!flag) dp[pos][r][mp[lcm]]=best;
    return best;
}
ll solve(ll x)
{
    ll i,j,t,tot=0;
    while(x)
    {
        dig[++tot]=x%10;
        x/=10;
    }
   // printf("");
    t=dfs(tot,0,1,1);
    return t;
}
int main()
{
    ll i,j,t,le,ri,a2,a3,a5,a7;
    cnt=0;
    for(i=0;i<=3;i++)
    {
        if(i==0) a2=1;
        else a2*=2;
        for(j=0;j<=2;j++)
        {
            if(j==0) a3=1;
            else a3*=3;
            for(ll p=0;p<=1;p++)
            {
                if(p==0) a5=1;
                else a5*=5;
                for(ll q=0;q<=1;q++)
                {
                    if(q==0) a7=1;
                    else a7*=7;
                    mp[a2*a3*a5*a7]=++cnt;
                }
                a7=1;
            }
            a5=1;
        }
        a3=1;
    }
    fac[0]=1;
    for(i=1;i<=20;i++)
    {
        fac[i]=fac[i-1]*10;
    }
    memset(dp,-1,sizeof(dp));
    cin>>t;
    while(t--)
    {
        cin>>le>>ri;
        ans=solve(ri)-solve(le-1);
        cout<<ans<<endl;
    }
    return 0;
}