IT City company developing computer games invented a new way to reward its employees. 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 divisible by all numbers 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.
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 one integer showing how many numbers from 1 to n are divisible by all numbers from 2 to 10.
3000
1
题意：给你一个长整形数n，让你计算出1到n中有多少个数可以被2到10之间的所有数整除，包括2和10；
题解：先打表找规律，发现任意两个相邻的满足条件的两个数之间相差 2520，所以拿n除以2520就是结果
#include<stdio.h> //j
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#define INF 0x3f3f3f
#define MAX 100100
#define LL long long
using namespace std;
int main()
{
LL n,m,j,i;
LL sum;
while(scanf("%lld",&n)!=EOF)
{
sum=n/2520;
printf("%lld
",sum);
// sum=0; //打表找规律
// for(i=1;i<=n;i++)
// {
// int flag=1;
// for(j=6;j<=10;j++)
// {
// if(i%j!=0)
// {
// flag=0;
// break;
// }
// }
// if(flag)
// {
// printf("%d ",i);
// sum++;
// }
// }
// printf("
%d
",sum);
}
return 0;
}