Codeforces--630J--Divisibility（公倍数）

最新推荐文章于 2022-10-20 17:20:36 发布

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最新推荐文章于 2022-10-20 17:20:36 发布

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原文链接：http://www.cnblogs.com/playboy307/p/5273413.html

本文介绍了一个简单的算法，用于计算一款游戏在首次发布的前一个月内，根据预计销量获得奖金的次数。通过找出2到10之间的最小公倍数，即2520，文章中的C++代码示例展示了如何快速计算出满足条件的销售数量。

J - Divisibility

Crawling in process... Crawling failed Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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.

Sample Input

Input

Output

求在1--n中可以被2--10全部整除的数字的个数，很明显，被2--10全部整除就是求2--10这几个数的最小公倍数，好吧，就是2520

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
	__int64 n;
	while(cin>>n)
	{
		cout<<n/2520<<endl;
	}
	return 0;
}

转载于:https://www.cnblogs.com/playboy307/p/5273413.html