題目思路：數位dp，若這個數能被每位的非0數整除，那麽這個數一定可以被每一位數的lcm整除，lcm（1,2,3,4,5,6,7,8,9） = 2520，所以可以通過將這個數對2520取模來壓縮空間,取模結果計做mod

dp[pos][lcm][mod]，顯然20*2520*2520仍然過大，所以我們對mod進行離散，再次壓縮空間。

#include <iostream>
#include <cmath>
#include <string.h>
#include <cstdio>
#include <queue>
#include 
 
<algorithm>
using namespace std;
#define LL long long
#define MAXSIZE 2600
#define INF 0x3f3f3f3f

LL dp[20][50][MAXSIZE];
const LL Mod = 2520;

int bit[MAXSIZE],ha[MAXSIZE];

LL gcd(LL n,LL m)
{
    if(n%m == 0)
        return m;
    return gcd(m,n%m);
}

void Init()
{
    int pos = 1
 
;
    for(int i=1;i<=Mod;i++)
    {
        if(2520%i == 0)
            ha[i] = pos++;
    }
}

LL dfs(int pos,int lcm,int mod,int limit)
{
    if(pos < 0)
    {
        if(mod%lcm == 0)
            return 1;
        return 0;
    }

    if(dp[pos][ha[lcm]][mod] != -1 && !limit)
        
 
return dp[pos][ha[lcm]][mod];

    LL ans = 0;
    int len = limit?bit[pos]:9;
    for(int i=0;i<=len;i++)
    {
        if(i != 0)
        {
            int new_lcm = lcm*i/(gcd(lcm,i));
            int num = (mod*10+i)%Mod;
            ans += dfs(pos-1,new_lcm,num,limit&&i==len);
        }

        else
        {
            int num = mod*10%Mod;
            ans += dfs(pos-1,lcm,num,limit&&i==len);
        }
    }
    if(!limit)
        dp[pos][ha[lcm]][mod] = ans;
    return ans;
}

LL Solve(LL n)
{
    int pos = 0;
    while(n)
    {
        bit[pos++] = n%10;
        n /= 10;
    }
    LL ans = dfs(pos-1,1,0,1);
    return ans;
}

int main()
{
    Init();
    int T;
    LL n,m;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        LL ans = Solve(m) - Solve(n-1);
        printf("%lld\n",ans);
    }
    return 0;
}

51nod 1232 完美數