codeforces 630K Indivisibility （容斥原理）

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

题意：给你一个数n，找出1~n范围内不被2~10整除的数的个数。

思路：这题可以用容斥原理，找到2~10里的4个素数2,3,5,7,然后用容斥原理就行了。

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<string>

#include<algorithm>

using namespace std;

typedef long long ll;

#define inf 99999999

#define pi acos(-1.0)

int main()

{

    ll n,ans;

    while(scanf("%I64d",&n)!=EOF)

    {

        ans=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("%I64d\n",ans);

    }

    return 0 ;

}

巴特西

codeforces 630K Indivisibility （容斥原理）

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