codeforces 630KIndivisibility(容斥原理)
2016-03-10 20:23
477 查看
IndivisibilityDescriptionIT 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 gameduring the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.InputThe only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the numberof people who will buy the game.OutputOutput one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.Sample InputInput
12Output
2
题意:求出1到n中不能被2到10中任意一个数整除的数的个数。
思路:2的倍数有2,4,6,8,10,3的倍数有3,6,9,因为能被整除2,4,6,8,10的数一定能被2 整除,能被整除3,6,9的数一定能被3 整除,所以只要求出不能被2,3,5,7整除的数即可,用容斥原理可得,n-m(m指能被2,3,5,7整除的数),则m等于2,3,5,7的倍数和减去两两重叠的加上三个三个重叠的减去四个重叠的。
代码:
#include<stdio.h>int main(){long long n,m;while(scanf("%lld",&n)!=EOF){m=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-m);}return 0;}
相关文章推荐
- clang-format中文出错
- poj 2407 Relatives
- 第2周项目3 小试循环
- 单例模式
- LeetCode 8. String to Integer (atoi)
- Leetcode 160 Intersection of Two Linked Lists 单向链表
- 一个return引发的血案
- Python基础——module
- 客户端——解析json数据
- BigInteger类的一些用法
- android开发之java内存泄露分析
- 从angularJS改道Vue.js,趟过第一个坑!
- JS里的onclick事件
- Pg188-3 构造方法
- BZOJ 1264: [AHOI2006]基因匹配Match 树状数组+DP
- 携程Android App插件化和动态加载实践
- Nginx为什么比Apache Httpd高效:原理篇
- content-type 类型讲解
- Apache commons-io
- 跳青蛙问题与变态跳青蛙问题