链接:

https://codeforces.com/contest/1209/problem/A

题意:

You are given a sequence of integers a1,a2,…,an. You need to paint elements in colors, so that:

If we consider any color, all elements of this color must be divisible by the minimal element of this color.

The number of used colors must be minimized.

For example, it's fine to paint elements [40,10,60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.

For example, if a=[6,2,3,4,12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10,7,15] then 3 colors are required (we can simply paint each element in an unique color).

思路:

暴力枚举.

代码:

#include <bits/stdc++.h>
using namespace std; int a[110];
int vis[110]; int main()
{
int n;
cin >> n;
for (int i = 1;i <= n;i++)
cin >> a[i];
sort(a+1, a+1+n);
int col = 0;
for (int i = 1;i <= n;i++)
{
if (vis[i] == 0)
{
col++;
for (int j = i;j <= n;j++)
{
if (a[j]%a[i] == 0)
vis[j] = 1;
}
}
}
cout << col << endl; return 0;
}

Codeforces Round #584 A. Paint the Numbers的更多相关文章

  1. Codeforces Round #584 C. Paint the Digits

    链接: https://codeforces.com/contest/1209/problem/C 题意: You are given a sequence of n digits d1d2-dn. ...

  2. Codeforces Round #584

    传送门 A. Paint the Numbers 签到. Code #include <bits/stdc++.h> using namespace std; typedef long l ...

  3. Codeforces Round #584 - Dasha Code Championship - Elimination Round (rated, open for everyone, Div. 1 + Div. 2)

    怎么老是垫底啊. 不高兴. 似乎 A 掉一道题总比别人慢一些. A. Paint the Numbers 贪心,从小到大枚举,如果没有被涂色,就新增一个颜色把自己和倍数都涂上. #include< ...

  4. Codeforces Round #131 (Div. 1) B. Numbers dp

    题目链接: http://codeforces.com/problemset/problem/213/B B. Numbers time limit per test 2 secondsmemory ...

  5. Educational Codeforces Round 13 A. Johny Likes Numbers 水题

    A. Johny Likes Numbers 题目连接: http://www.codeforces.com/contest/678/problem/A Description Johny likes ...

  6. Educational Codeforces Round 23 C. Really Big Numbers 暴力

    C. Really Big Numbers time limit per test 1 second memory limit per test 256 megabytes input standar ...

  7. Codeforces Round #584 E2. Rotate Columns (hard version)

    链接: https://codeforces.com/contest/1209/problem/E2 题意: This is a harder version of the problem. The ...

  8. Codeforces Round #584 D. Cow and Snacks

    链接: https://codeforces.com/contest/1209/problem/D 题意: The legendary Farmer John is throwing a huge p ...

  9. Codeforces Round #584 B. Koala and Lights

    链接: https://codeforces.com/contest/1209/problem/B 题意: It is a holiday season, and Koala is decoratin ...

随机推荐

  1. [转帖]hive与hbase的联系与区别:

    https://www.cnblogs.com/xubiao/p/5571176.html 原作者写的很好.. 这里面简单学习总结一下.. 都是bigdata的工具, 都是基于google的bigta ...

  2. 龙芯PG10 安装uuid-ossp 的方法 复用瀚高数据库的 so文件

    接着上一篇blog  当时在中标麒麟 龙芯上面安装了postgresql10.10 的版本 但是没搞定 uuid 当时遇到的问题: 0. 只安装postgresql数据库会报错如图示: 我验证了下 安 ...

  3. VLAN之间通信-三层交换机实现

    1.打开三层交换机的命令行,配置VLAN和设置端口IP enable //进入特权模式 configure terminal //进入全局配置模式 ip routing //启动交换机的路由功能 vl ...

  4. gdb 常用命令总结(精优)

    格式说明: [xxx]:可选参数,即可以指定可以不指定,实际输入的内容是 xxx <xxx>:占位参数,即必须指定的参数,实际输入的内容是 xxx gdb 常用命令: gdb [file] ...

  5. C++:标准模板库Sort

    一.概述 STL几乎封装了所用的数据结构中的算法,这里主要介绍排序算法的使用,指定排序迭代器区间后,即可实现排序功能. 所需头文件#include <algorithm> sort函数:对 ...

  6. pipreqs 生成项目依赖的第三方包

    项目开发的时候,总是要搭建和部署环境. 如果项目使用virtualenv环境,直接使用使用命令行pip freeze可以帮助我们自动生成项目所需要的环境 requirements.txt文件 $ pi ...

  7. 从业务流程角度:分析TMS系统各个功能模块

    TMS的主要功能是协调承运商.运营商.货主三种角色人员分工合作共同完成运输任务,并实现对运输任务的跟踪管理.本文将按照业务流程顺序对TMS系统各个功能模块进行分析说明. 一.业务描述 新零售的兴起及& ...

  8. javascript 正则表达式的简单操作

    前言:这是笔者学习之后自己的理解与整理.如果有错误或者疑问的地方,请大家指正,我会持续更新! RegExp 正则表达式是描述字符模式的对象. 正则表达式用于对字符串模式匹配及检索替换,是对字符串执行模 ...

  9. 浅谈C++继承

    C++中的继承 1.继承概念及定义:     概念:是面向对象程序设计使代码可以复用的最重要的手段-----继承是类设计层次的复用     定义:            父类->基类:子类-&g ...

  10. 区间dp最长回文子序列问题

    状态转移方程如下: 当i > j时,dp[i,j]= 0. 当i = j时,dp[i,j] = 1. 当i < j并且str[i] == str[j]时,dp[i][j] = dp[i+1 ...