cf55dBeautiful numbers数位dp

想到 最小公倍数 其余的就好搞了 ，可是没想到

#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;

}