数位dp整理

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxa = 20;
const int mod = 2520;
long long  dp[maxa][1<<8][mod];
#define LL long long
//当前位置，限制，有哪些数，模
int num[maxa];
LL dfs(int pos, int limit, int val, int numb){
    if(pos == 0){
        for(int i = 2; i < 10; i++){
            if((1<<(i-2)) & val){
                if(numb % i) return 0;
            }
        }return 1;
    }
    if(!limit && dp[pos][val][numb] != -1) return dp[pos][val][numb];
    int nn = limit?num[pos]:9;
    LL res = 0;
    for(int i = 0; i <= nn; i++){
        res += dfs(pos-1, limit & (i ==nn), i<2?val:val|(1<<(i-2)), (numb*10+i)%mod);
    }
    if(!limit) dp[pos][val][numb] = res;
    return res;
}
LL ANS(LL n){
    int leng = 0;
    while(n){
        num[++leng] = n %10;
        n /= 10;
    }
    return dfs(leng, 1, 0, 0);
}
int main(){
    int t;
    LL a, b;
    memset(dp, -1, sizeof(dp));
    scanf("%d", &t);
    while(t--){
        cin>>a>>b;
        if(a > b)swap(a, b);
        cout<<ANS(b) - ANS(a-1)<<endl;
    }
}