cf55dBeautiful numbers数位dp

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;
typedef long long LL;


const int  mod=2520;
LL  dp[20][2520][1<<8];
int  up[20];
LL dfs(int now,int pre,int gaojici,int flag)
{
    if(now<=0){
        for(int i=0;i<=7;i++)
        if((gaojici&(1<<i))&&pre%(i+2)) return 0;
        return 1;
    }
    if(!flag&&~dp[now][pre][gaojici]) return dp[now][pre][gaojici];
    int limit = flag? up[now] : 9;LL ret=0 ;
    for(int  i = 0;i<=limit;i++){
        if(i>=2)
        ret+=dfs(now-1,(pre*10+i)%mod, gaojici|(1<<(i-2)) ,flag&&i==limit);
        else
        ret+=dfs(now-1,(pre*10+i)%mod, gaojici,flag&&i==limit);
    }
    return flag? ret: dp[now][pre][gaojici]= ret;
}
LL solve(LL  x)
{
    int len =0 ;
    while(x){
        up[++len] = x%10;
        x/=10;
    }
    return dfs(len ,0, 0 , 1)-1;
}
int   main()
{
    int  Icase;LL  a,b;
    scanf("%d",&Icase);
    memset(dp,-1,sizeof(dp));
    for(int i = 1;i<=Icase;i++){
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",solve(b) - solve(a-1));
    }
    return 0;
}