[暑假集训--数位dp]cf55D Beautiful numbers

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 9 ·10¹⁸).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from l_i to r_i, inclusively).

Examples

Input

1
1 9

Output

Input

1
12 15

Output

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<LL,LL>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
23     return x*f;
24 }
25 LL n,len,l,r,tot;
26 LL f[20][210][2530];
27 LL d[110];
28 LL rnk[2530];
29 LL mrk[210];
30 inline LL dfs(LL now,LL lcm,LL rem,LL fp)
31 {
32     if (now==1)return rem%lcm==0;
33     if (!fp&&f[now][rnk[lcm]][rem]!=-1)return f[now][rnk[lcm]][rem];
34     LL ans=0;
35     LL mx=fp?d[now-1]:9;
36     LL nexlcm,nexrem=rem*10%2520;
37     for (LL i=0;i<=mx;i++)
38     {
39         nexlcm=lcm;
40         if (i)nexlcm=nexlcm*i/__gcd(lcm,i);
41 
42         nexrem+=i;if (nexrem>=2520)nexrem-=2520;
43 
44         ans+=dfs(now-1,nexlcm,nexrem,fp&&i==mx);
45 
46         nexrem-=i;if (nexrem<0)nexrem+=2520;
47     }
48     if (!fp)f[now][rnk[lcm]][rem]=ans;
49     return ans;
50 }
51 inline LL calc(LL x)
52 {
53     if (x==-1)return 0;
54     if (x==0)return 1;
55     LL xxx=x;
56     len=0;
57     while (xxx)
58     {
59         d[++len]=xxx%10;
60         xxx/=10;
61     }
62     LL sum=0;
63     for (LL i=0;i<=d[len];i++)
64     {
65         sum+=dfs(len,i?i:1,i,i==d[len]);
66     }
67     return sum;
68 }
69 inline void init()
70 {
71     priority_queue<LL,vector<LL>,greater<LL> >q;
72     q.push(1);
73     while (1)
74     {
75         LL now=q.top();q.pop();
76         if (now>2520)break;
77         if (rnk[now])continue;
78         for (LL i=2;i<=10;i++)q.push(i*now);
79         rnk[now]=++tot;
80         mrk[tot]=now;
81     }
82 }
83 int main()
84 {
85     init();
86     LL T=read();
87     memset(f,-1,sizeof(f));
88     while (T--)
89     {
90         l=read();
91         r=read();
92         if (r<l)swap(l,r);
93         printf("%lld\n",calc(r)-calc(l-1));
94     }
95 }

cf 55D

posted on 2017-08-04 12:20 zhber 阅读(177) 评论(0) 编辑收藏举报