【CodeForces 630K】Indivisibility(容斥原理)

K. Indivisibility

time limit per test

0.5 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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

                                                            

题意及思路：

求1-n中不能被2-10中任何数整除的数的个数。如果直接遍历求解，肯定会超时。后来发现能被4，6，8，10整除的一定被2整除，同理被6，9整除的一定被3整除，被10整除的一定被5整除，因此，2-10中只需要判断2，3，5，7即可。但是这样还不够，当数据很大时依旧会超时。因此可以想到容斥原理来解此题。由于减去2、3、5、7的倍数时，会重复减去一些数（如6、10、14），所以，最后再加上两数之积，减去三数之积，加上四数之积即可。

#include<stdio.h>int main(){long long int a,ans;while(scanf("%I64d",&a)!=EOF){ans=a-(a/2+a/3+a/5+a/7)+(a/6+a/10+a/14+a/15+a/21+a/35)-(a/30+a/42+a/70+a/105)+a/210;printf("%I64d\n",ans);}return 0;}