Codeforces 630J Divisibility

最新推荐文章于 2020-01-29 23:14:26 发布

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最新推荐文章于 2020-01-29 23:14:26 发布

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本文介绍了一个基于游戏销售数量预测员工奖金的方法。每当游戏销量能够被2到10之间的所有数字整除时，开发者就能获得一次奖金。文章通过一个具体的例子说明了如何计算这种情况下开发者可以获得奖金的次数。

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

求从1到n有多少个数可以被从2到10的数整除，最小公倍数

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;

int main()
{
    long long n,sum;
    while(~scanf("%I64d",&n))
    {
        printf("%I64d\n",n/2520);
    }
    return 0;
}