CF #603K Indivisibility

最新推荐文章于 2023-10-07 19:58:43 发布

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最新推荐文章于 2023-10-07 19:58:43 发布

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分类专栏： ACM 文章标签： acm模板

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本文介绍了一个游戏开发公司根据游戏销量给开发者发放奖金的算法。利用容斥原理计算1至n间不可被2至10中任一数字整除的数字数量。

/又学习到新技能，貌似大二会学，不过提前了解了一点/
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 thatn 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 ton are not divisible by any number from2 to 10.

Example
Input
12
Output
2

/*题目大意就是：找出1–n之间不能被2–10之间任意数整除的数的个数
利用容斥原理解决问题可以事半功倍，2-10之间可看做2,3,5,7，这四个数，因为4,6,8,9,10是他们的倍数，*/

#include<cstdio>
int main()
{
    long long n,sum;
    scanf("%lld",&n);
    sum=n-(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",sum);
    return 0;
}