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 li and ri (1 ≤ li ≤ ri ≤ 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 li to ri, inclusively).

Examples
Input
1
1 9
Output
9
Input
1
12 15
Output
2

  

 #include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<LL,LL>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
LL n,len,l,r,tot;
LL f[][][];
LL d[];
LL rnk[];
LL mrk[];
inline LL dfs(LL now,LL lcm,LL rem,LL fp)
{
if (now==)return rem%lcm==;
if (!fp&&f[now][rnk[lcm]][rem]!=-)return f[now][rnk[lcm]][rem];
LL ans=;
LL mx=fp?d[now-]:;
LL nexlcm,nexrem=rem*%;
for (LL i=;i<=mx;i++)
{
nexlcm=lcm;
if (i)nexlcm=nexlcm*i/__gcd(lcm,i); nexrem+=i;if (nexrem>=)nexrem-=; ans+=dfs(now-,nexlcm,nexrem,fp&&i==mx); nexrem-=i;if (nexrem<)nexrem+=;
}
if (!fp)f[now][rnk[lcm]][rem]=ans;
return ans;
}
inline LL calc(LL x)
{
if (x==-)return ;
if (x==)return ;
LL xxx=x;
len=;
while (xxx)
{
d[++len]=xxx%;
xxx/=;
}
LL sum=;
for (LL i=;i<=d[len];i++)
{
sum+=dfs(len,i?i:,i,i==d[len]);
}
return sum;
}
inline void init()
{
priority_queue<LL,vector<LL>,greater<LL> >q;
q.push();
while ()
{
LL now=q.top();q.pop();
if (now>)break;
if (rnk[now])continue;
for (LL i=;i<=;i++)q.push(i*now);
rnk[now]=++tot;
mrk[tot]=now;
}
}
int main()
{
init();
LL T=read();
memset(f,-,sizeof(f));
while (T--)
{
l=read();
r=read();
if (r<l)swap(l,r);
printf("%lld\n",calc(r)-calc(l-));
}
}

cf 55D

[暑假集训--数位dp]cf55D Beautiful numbers的更多相关文章

  1. [暑假集训--数位dp]LightOj1205 Palindromic Numbers

    A palindromic number or numeral palindrome is a 'symmetrical' number like 16461 that remains the sam ...

  2. 【数位dp】Beautiful Numbers @2018acm上海大都会赛J

    目录 Beautiful Numbers PROBLEM 题目描述 输入描述: 输出描述: 输入 输出 MEANING SOLUTION CODE Beautiful Numbers PROBLEM ...

  3. [暑假集训--数位dp]hdu3709 Balanced Number

    A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. ...

  4. [暑假集训--数位dp]hdu3555 Bomb

    The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the ti ...

  5. [暑假集训--数位dp]hdu3652 B-number

    A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- ...

  6. [暑假集训--数位dp]hdu2089 不要62

    杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer).杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以消除个别的士司机和乘客的心理障碍, ...

  7. [暑假集训--数位dp]hdu5787 K-wolf Number

    Alice thinks an integer x is a K-wolf number, if every K adjacent digits in decimal representation o ...

  8. [暑假集训--数位dp]LightOJ1140 How Many Zeroes?

    Jimmy writes down the decimal representations of all natural numbers between and including m and n, ...

  9. [暑假集训--数位dp]LightOj1032 Fast Bit Calculations

    A bit is a binary digit, taking a logical value of either 1 or 0 (also referred to as "true&quo ...

随机推荐

  1. leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal,剑指offer 6 重建二叉树

    不用迭代器的代码 class Solution { public: TreeNode* reConstructBinaryTree(vector<int> pre,vector<in ...

  2. Python -- 函数之推导式

    5.12 推导式 l = [] for i in range(1,11): l.append(i) print(l) # 用列表推导式 (一行搞定) l = [i for i in range(1,1 ...

  3. [vijos]P1979 NOIP2015 信息传递

    描述 有 n 个同学(编号为 1 到 n)正在玩一个信息传递的游戏.在游戏里每人都有一个固定的信息传递对象,其中,编号为 i 的同学的信息传递对象是编号为 TiTi 的同学. 游戏开始时,每人都只知道 ...

  4. 洛谷P2347 砝码称重

    题目 貌似是某年提高组签到题,六重循环零压力AC,差点怒踩std 但本蒟蒻决定写正解——多重背包,果断20分 原因是写错了状态转移方程...神才知道我咋过的样例和两个测试点 扯远了 多重背包 简单说一 ...

  5. python列表之append与extend方法比较

    append和extend是列表的两种添加元素方式,但这两种方式却又有些不同之处.那么不同之处在哪里呢,我们通过对二者的定义和实例来看一看. list.append() 1.定义:L.append(o ...

  6. Kubernetes(k8s)底层网络原理刨析

    目录 1 典型的数据传输流程图 2 3种ip说明 3 Docker0网桥和flannel网络方案 4 Service和DNS 4.1 service 4.2 DNS 5 外部访问集群 5.1 外部访问 ...

  7. 水题:UVa133-The Dole Queue

    The Dole Queue Time limit 3000 ms Description In a serious attempt to downsize (reduce) the dole que ...

  8. cf 1016C

    C. Vasya And The Mushrooms time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  9. UML结构与解析——BUAA OO第四单元作业总结

    UML与解析架构 UML是什么 统一建模语言(英语:Unified Modeling Language,缩写 UML)是非专利的第三代建模和规约语言.UML是一种开放的方法,用于说明.可视化.构建和编 ...

  10. Nginx与Lua的开发

    1. Lua基础语法 安装lua hello world 也可以编写lua脚本 运行脚本 lua注释 变量 局部变量的话前面加个local 循环 if语句 2. Nginx与Lua开发环境 https ...