CodeForces 630K Indivisibility(容斥原理)

Indivisibility
Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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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
12
Output
2

分析：
容斥原理。2/4/6/8/10 公约数为2, 6/9 公约数为3, 2~10中还剩两个5, 7.
最后只要考虑2,3,5,7的倍数。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll ans,n;
int main(){
    cin>>n;
    ll tmp=n-n/2-n/3-n/5-n/7 +n/6+n/10+n/14+n/15+n/21+n/35;
    tmp=tmp-n/30-n/42-n/70-n/105 +n/210;
    cout<<tmp<<endl;
    return 0;
}