#include<string> #include <string.h> #include <stdio.h> #include <stdlib.h> #include <math.h> #include <algorithm> #include <queue> #include <vector> #include <cstdio> #include <iostream> using namespace std; typedef long long ll; const int maxn=2550; //原数是不是整除非零位,也就是整除最小公倍数,1-9所有最小公倍数为2520 int a[20],mp[maxn];//1-9任意最小公倍数是48个 ll d[20][maxn][50];//第几位,取模后为sum,最小公倍数为lcm的个数 ll n,m; void init() { int ans=0; for(int i=1; i<=2520; i++) if(2520%i==0) { mp[i]=ans; ans++; } } ll work(ll x,ll y) { if(x==0||y==0) return x+y; else return x*y/__gcd(x,y); } ll dfs(int pos,ll sum,int lcm,bool limit) { if(pos==-1) { return (sum%lcm)==0; } if(!limit&&d[pos][sum][mp[lcm]]!=-1) return d[pos][sum][mp[lcm]]; int u=limit?a[pos]:9; ll temp=0; for(int i=0; i<=u; i++) { temp+=dfs(pos-1,(sum*10+i)%2520,work(lcm,i),limit&&a[pos]==i); } if(!limit) d[pos][sum][mp[lcm]]=temp; return temp; } ll solve(ll x) { int pos=0; while(x>0) { a[pos++]=int(x%10); x/=10; } return dfs(pos-1,0,1,true);//必须为1,否则sum%lcm报错 } int main() { memset(d,-1,sizeof d); init(); int t; cin>>t; while(t--) { cin>>n>>m; cout<<(solve(m)-solve(n-1))<<endl; } return 0; }