Experimental Educational Round: VolBIT Formulas Blitz K. Indivisibility —— 容斥原理

最新推荐文章于 2022-07-16 16:36:08 发布

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最新推荐文章于 2022-07-16 16:36:08 发布

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分类专栏：组合数学

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本文介绍了一道Codeforces上的编程题——K.Indivisibility的解题思路与实现代码。该题要求计算从1到n中不能被2至10之间任意数整除的整数数量，采用容斥原理解决。

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题目链接：http://codeforces.com/contest/630/problem/K

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
   
12

     output
   
2

题解：

实际上是除去2,3,5,7的倍数，容斥原理。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 1e3+10;

int main()
{
    LL n;
    cin>>n;
    LL ans = 0, a[4] = {2,3,5,7};
    for(int s = 1; s<(1<<4); s++)
    {
        LL cnt = 0, val = 1;
        for(int j = 0; j<4; j++)
        if((1<<j)&s)
        {
            cnt++;
            val *= a[j];
        }
        ans += (cnt&1)? (n/val):(-n/val);
    }
    cout << n-ans <<endl;
}