codeforces 630KIndivisibility（容斥原理）

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最新推荐文章于 2022-11-07 15:01:58 发布

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本文讨论了IT城游戏公司奖励机制的实现方式，通过跟踪新游戏的销售情况，并根据销售数量是否能被2到10之间的任意数整除来决定开发者是否获得奖金。设计师Petya预测了新游戏首月的销量并希望了解自己将获得多少次奖金。

Indivisibility

Description

IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. 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 not divisible by any number 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 not divisible by any number from 2 to 10.

Sample Input

Input

Output

题意：求出1到n中不能被2到10中任意一个数整除的数的个数。

思路：2的倍数有2，4，6，8，10，3的倍数有3，6，9，因为能被整除2，4，6，8，10的数一定能被2 整除，能被整除3，6，9的数一定能被3 整除，所以只要求出不能被2，3，5，7整除的数即可，用容斥原理可得，n-m（m指能被2，3，5，7整除的数），则m等于2，3，5，7的倍数和减去两两重叠的加上三个三个重叠的减去四个重叠的。

代码：

#include<stdio.h>
int main()
{
	long long n,m;
	while(scanf("%lld",&n)!=EOF)
	{
		m=n/2+n/3+n/5+n/7-n/6-n/10-n/14-n/15-n/21-n/35+n/30+n/42+n/70+n/105-n/210;
		printf("%lld\n",n-m);
	}
	return 0;
}