CodeForces-630 J. Divisibility

最新推荐文章于 2020-01-29 23:14:26 发布

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最新推荐文章于 2020-01-29 23:14:26 发布

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本文探讨了一款游戏发行后，通过跟踪销售数据，如何精确计算每达到特定销售数量区间时，开发者能获得奖金的次数。特别关注了如何利用数学原理解决实际业务问题。

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.

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 divisible by all numbers from 2 to 10.

Examples

input
3000

output
1

题意：

问1-n中有多少个能被2-10的所有数整除

题解：

求出2-10的最小公倍数x，就是求1--n中多少个数能被x整除了...

#include<stdio.h>
typedef long long ll;
ll gcd(ll a,ll b)
{
    if(!b)
    {
        return a;
    }
    return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    ll n,tp=2;
    for(ll i=3;i<=10;++i)
    {
        tp=lcm(tp,i);
    }
    while(~scanf("%lld",&n))
    {
        printf("%I64d\n",n/tp);
    }
    return 0;
}