Codeforces Beta Round #51 D. Beautiful numbers

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 li and ri (1 ≤ li ≤ ri ≤ 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 li to ri, inclusively).
Sample test(s)
input

1

1 9

output

9

input

1

12 15

output

2

  dp,  ，      ，  ，         ，  ，       ，                    ！     dp[i][j][]   i ，      j， i      k       ，         ，          ，  ，２－９        ２５２０，    ，  ４８ ，  ，       ，    ！

#include <iostream>

#include <stdio.h>

#include <string.h>

using namespace std;

#define M 23

__int64 dp[M][55][3500],pri[M];

int hash[3500],kk=0;

int gcd(int a,int b){

    if(a==0)return b;

    return gcd(b%a,a);

}

int get(int t,int i){

    if(i==0||i==1)return t;

    return t*i/gcd(t,i);

}

__int64 dfs(int pos,int t,int flag,int pre){

    if(pos==0)return pre%t==0;

    if(!flag&&dp[pos][hash[t]][pre]!=-1)return dp[pos][hash[t]][pre];

    int u=flag?pri[pos]:9;

    __int64 ans=0;

    for(int i=0;i<=u;i++)

    ans+=dfs(pos-1,get(t,i),flag&&i==u,(pre*10+i)%2520);

    return flag?ans:dp[pos][hash[t]][pre]=ans;

}

__int64 solve(__int64 x){

    int cnt=0;

    while(x){

        pri[++cnt]=x%10;x/=10;

    }

    return dfs(cnt,1,1,0);

}

int init(){

    memset(dp,-1,sizeof(dp));

    int cnt=0;

    hash[0]=0;

    for(int i=1;i<=8;i*=2)

        for(int j=1;j<=9;j*=3)

            for(int k=1;k<=5;k*=5)

                for(int x=1;x<=7;x*=7)

                hash[i*j*k*x]=++cnt;

}

int main()

{

    int tcase;__int64 n,m;

    init();

    while(scanf("%d",&tcase)!=EOF){

        while(tcase--){

            scanf("%I64d%I64d",&m,&n);

            printf("%I64d
",solve(n)-solve(m-1));

        }

    }

    return 0;

}