cf 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 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

求[l,r]中能整除各个位数（0除外）的数的个数。

CODE：

#include  
#include  
#include  
#include  
#include  
#define ll long long  
#define Mod 2520  ////1-9的最小公倍数  
  
using namespace std;  
  
ll l,r;  
ll dp[30][Mod][50];///dp[i][j][k] i:长度  j:各个位数最小公倍数对Mod取摸  k:最小公倍数对应的值得离散下标  
int num[30];  
int b[Mod+1];     ///记录最小公倍数  
  
ll gcd(ll a,ll b) {  
    return b==0?a:gcd(b,a%b);  
}  
  
ll Lcm(ll a,ll b) {  
    return a/gcd(a,b)*b;  
}  
  
ll dfs(int i,ll lcm,ll sum,bool e) {  
    if(i<=0)return sum%lcm==0;  
    if(!e&&dp[i][sum][b[lcm]]!=-1)return dp[i][sum][b[lcm]];  
    ll res=0;  
    int u=e?num[i]:9;  
    for(ll d=0; d<=u; d++) {  
        if(d==0)                 ///d==0时，lcm不变  
            res+=dfs(i-1,lcm,sum*10%Mod,e&&d==u);  
        else  
            res+=dfs(i-1,lcm*d/gcd(lcm,d),(sum*10+d)%Mod,e&&d==u);  
    }  
    return e?res:dp[i][sum][b[lcm]]=res;  
}  
   
ll solve(ll n) {  
    int len=1;  
    while(n) {  
        num[len++]=n%10;  
        n/=10;  
    }  
    return dfs(len-1,1,0,1);  
}  
  
int main() {  
    memset(dp,-1,sizeof dp);  
    int cnt=0;  
[cpp] view plaincopy
    ///离散记录lcm值  
    for(int i=1; i<=Mod; i++)if(Mod%i==0)b[i]=cnt++;  
    int t;  
    scanf("%d",&t);  
    while(t--) {  
        scanf("%I64d%I64d",&l,&r);  
        ll ans=solve(r)-solve(l-1);  
        printf("%I64d\n",ans);  
    }  
    return 0;