Experimental Educational Round: VolBIT Formulas Blitz K

最新推荐文章于 2025-04-23 11:34:14 发布

转载最新推荐文章于 2025-04-23 11:34:14 发布 · 292 阅读

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原文链接：http://www.cnblogs.com/yinghualuowu/p/5228324.html

本文介绍了一家游戏公司的员工奖励机制，每当游戏销售数量不能被2到10之间的任何数整除时，开发者将获得小额度奖金。文章通过具体实例和代码详细解释了如何计算在预测销量下，开发者能获得奖金的次数。

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.

Examples

input

output

2
容斥原理

#include<iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
typedef long long LL;
int num[20];
int n;
LL sum;//n是num[]的元素个数
LL m;
LL gcd(LL a,LL b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
LL lcm(LL a,LL b)
{
    return a*b/gcd(a,b); //这是求最小公倍数的方法
}
LL dfs(LL lcmn,int id)    //这里传进来的lcmn是long long !!!
{
    if(id<n-1) return dfs(lcmn,id+1)-dfs(lcm(lcmn,num[id+1]),id+1);
    return m/lcmn;
}
int main()
{
    int t;
    n=9;
    cin>>m;
    for(int i=0; i<n; i++)
    {
        num[i]=i+2;
    }
    sort(num,num+n);
    sum=0;
    for(int i=0; i<n; i++)
        sum+=dfs(num[i],i);
    cout<<m-sum<<endl;
    return 0;
}

转载于:https://www.cnblogs.com/yinghualuowu/p/5228324.html