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

数位dp,这题，要求最后的数，对于，每一个数字都要整除，那么，我们就可以通过，一个数能整除所有数字的最小公倍数就可以了！所以我们用dp[i][j][]表示第i位，最小公倍数为j，第i位时余下的是k时的最后的个数，这题还可以进行优化，最是把最小公倍数编号，因为，２－９最小公倍数最大是２５２０，也就是说，只有４８个，这样，大大节省了内存，加快速度！

#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\n",solve(n)-solve(m-1));
        }
    }
    return 0;
}