CF630K：Indivisibility(容斥）

 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

解体思路:算出2的倍数,3的倍数,5的倍数,7的倍数,减去任意两个数的倍数,加上任意三个数的倍数,减去任意4个数的倍数

代码如下:

#include<stdio.h>
int main(){
	long long n,m,x,ans;
	while(scanf("%lld",&n)!=EOF){
	   ans=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",n-ans);
	}
	return 0;
}