Description

IT City company developing computer games invented a new way to reward its employees. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is divisible by all numbers from 2 to 10 every developer of this game gets a small bonus.

A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.

Output

Output one integer showing how many numbers from 1 to n are divisible by all numbers from 2 to 10.

Examples
input
3000
output
1

问能同时被2到10整除的,有多少个

算出2到10的最小公倍数

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x7fffffff
#define INF 0x7fffffffffffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
LL n;
int main()
{
cin>>n;
cout<<n/2520<<endl;
return 0;
}

  

Experimental Educational Round: VolBIT Formulas Blitz J的更多相关文章

  1. Experimental Educational Round: VolBIT Formulas Blitz

    cf的一次数学场... 递推 C 题意:长度<=n的数只含有7或8的个数 分析:每一位都有2种可能,累加不同长度的方案数就是总方案数 组合 G 题意:将5个苹果和3个梨放进n个不同的盒子里的方案 ...

  2. Experimental Educational Round: VolBIT Formulas Blitz K. Indivisibility —— 容斥原理

    题目链接:http://codeforces.com/contest/630/problem/K K. Indivisibility time limit per test 0.5 seconds m ...

  3. Experimental Educational Round: VolBIT Formulas Blitz K

    Description IT City company developing computer games decided to upgrade its way to reward its emplo ...

  4. Experimental Educational Round: VolBIT Formulas Blitz N

    Description The Department of economic development of IT City created a model of city development ti ...

  5. Experimental Educational Round: VolBIT Formulas Blitz F

    Description One company of IT City decided to create a group of innovative developments consisting f ...

  6. Experimental Educational Round: VolBIT Formulas Blitz D

    Description After a probationary period in the game development company of IT City Petya was include ...

  7. Experimental Educational Round: VolBIT Formulas Blitz C

    Description The numbers of all offices in the new building of the Tax Office of IT City will have lu ...

  8. Experimental Educational Round: VolBIT Formulas Blitz B

    Description The city administration of IT City decided to fix up a symbol of scientific and technica ...

  9. Experimental Educational Round: VolBIT Formulas Blitz A

    Description The HR manager was disappointed again. The last applicant failed the interview the same ...

随机推荐

  1. spring bean属性及子元素使用总结

    spring bean属性及子元素使用总结 2016-08-03 00:00 97人阅读 评论(0) 收藏 举报  分类: Spring&SpringMVC(17)  版权声明:本文为博主原创 ...

  2. MySQL存储引擎 -- MyISAM 与 InnoDB 理论对比

    MySQL常用的两种存储引擎一个是MyISAM,另一个是InnoDB.两种存储引擎各有各的特点. 1. 区别:(1)事务处理:MyISAM是非事务安全型的.-----而非事务型的系统,一般也称为数据仓 ...

  3. jQuery选择器大全整理

    一.选择网页元素 $(document) //选择整个文档对象 $('#myId') //选择ID为myId的网页元素 $('div.myClass') // 选择class为myClass的div元 ...

  4. css田字格布局

    <!DOCTYPE html> <html> <head> <title></title> <style type="tex ...

  5. 内核文件ntoskrnl.exe,ntkrnlpa.exe的区别??

    除了标题中说到的两个exe文件之外,还有另外两个ntkrnlmp.exe和ntkrpamp.exe.因为我目前用到的只是标题中的两个. 其中,我在网上搜索到的关于SSDT HOOK 的资料,举的例子, ...

  6. java中public static void main(String[] args)中String[] args代表什么意思?

    这是java程序的入口地址,java虚拟机运行程序的时候首先找的就是main方法.跟C语言里面的main()函数的作用是一样的.只有有main()方法的java程序才能够被java虚拟机欲行,可理解为 ...

  7. bzoj4318 OSU!

    传送门 题目 osu 是一款群众喜闻乐见的休闲软件.  我们可以把osu的规则简化与改编成以下的样子:  一共有n次操作,每次操作只有成功与失败之分,成功对应1,失败对应0,n次操作对应为1个长度为n ...

  8. Luogu 3626 [APIO2009]会议中心

    很优美的解法. 推荐大佬博客 如果没有保证字典序最小这一个要求,这题就是一个水题了,但是要保证字典序最小,然后我就不会了…… 如果一条线段能放入一个区间$[l', r']$并且不影响最优答案,那么对于 ...

  9. beforeFilter()

    在控制器每个动作之前执行,可以方便地检查有效的会话,或者检查用户的权限. function beforeFilter() { parent::beforeFilter(); if(empty($thi ...

  10. MySQL中的时间问题

    MySQL 获得当前日期时间 函数 获得当前日期+时间(date + time)函数:now() mysql> select now(); +---------------------+ | n ...