Codeforces Round #603 (Div. 2) C. Everyone is a Winner! (数学)
链接:
https://codeforces.com/contest/1263/problem/C
题意:
On the well-known testing system MathForces, a draw of n rating units is arranged. The rating will be distributed according to the following algorithm: if k participants take part in this event, then the n rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.
For example, if n=5 and k=3, then each participant will recieve an 1 rating unit, and also 2 rating units will remain unused. If n=5, and k=6, then none of the participants will increase their rating.
Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.
For example, if n=5, then the answer is equal to the sequence 0,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋ for some positive integer k (where ⌊x⌋ is the value of x rounded down): 0=⌊5/7⌋, 1=⌊5/5⌋, 2=⌊5/2⌋, 5=⌊5/1⌋.
Write a program that, for a given n, finds a sequence of all possible rating increments.
思路:
枚举能得的分, n/sco 是sco对应的人数。再用n/人数,得到当前人数最大的分。
n/(n/num)下取整
代码:
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin >> t;
while(t--)
{
int n;
cin >> n;
int sco = 1;
vector<int> res;
res.push_back(0);
while(sco <= n)
{
int num = n/sco;
sco = n/num;
res.push_back(sco);
sco++;
}
cout << (int)res.size() << endl;
for (auto v: res)
cout << v << ' ' ;
cout << endl;
}
return 0;
}
Codeforces Round #603 (Div. 2) C. Everyone is a Winner! (数学)的更多相关文章
- Codeforces Round #603 (Div. 2) C. Everyone is a Winner! 二分
C. Everyone is a Winner! On the well-known testing system MathForces, a draw of n rating units is ar ...
- Codeforces Round #603 (Div. 2) C.Everyone is A Winner!
tag里有二分,非常的神奇,我用暴力做的,等下去看看二分的题解 但是那个数组的大小是我瞎开的,但是居然没有问题233 #include <cstdio> #include <cmat ...
- Codeforces Round #603 (Div. 2) A. Sweet Problem(水.......没做出来)+C题
Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory ...
- Codeforces Round #603 (Div. 2) E. Editor 线段树
E. Editor The development of a text editor is a hard problem. You need to implement an extra module ...
- Codeforces Round #603 (Div. 2) E. Editor(线段树)
链接: https://codeforces.com/contest/1263/problem/E 题意: The development of a text editor is a hard pro ...
- Codeforces Round #603 (Div. 2) D. Secret Passwords 并查集
D. Secret Passwords One unknown hacker wants to get the admin's password of AtForces testing system, ...
- Codeforces Round #603 (Div. 2) D. Secret Passwords(并查集)
链接: https://codeforces.com/contest/1263/problem/D 题意: One unknown hacker wants to get the admin's pa ...
- Codeforces Round #603 (Div. 2) B. PIN Codes
链接: https://codeforces.com/contest/1263/problem/B 题意: A PIN code is a string that consists of exactl ...
- Codeforces Round #603 (Div. 2) A. Sweet Problem(数学)
链接: https://codeforces.com/contest/1263/problem/A 题意: You have three piles of candies: red, green an ...
随机推荐
- opencv imshow plt imshow
opencv官方文档上写的,https://docs.opencv.org/master/dc/d2e/tutorial_py_image_display.html Color image loade ...
- 绝对有效 IntelliJ IDEA2019.2下载、安装及破解教程
原文链接:https://blog.csdn.net/weixin_43904316/article/details/88881238 https://bl ...
- 整理:WPF中Xaml中绑定枚举的写法
原文:整理:WPF中Xaml中绑定枚举的写法 目的:在Combobox.ListBox中直接绑定枚举对象的方式,比如:直接绑定字体类型.所有颜色等枚举类型非常方便 一.首先用ObjectDataPro ...
- WPF设置全局控件样式
原文:WPF设置全局控件样式 方法: 在资源文件APP.XAML中添加如下资源 <Application x:Class="_360UI.App" xmlns="h ...
- ASP.NET Core应用程序容器化、持续集成与Kubernetes集群部署(一)(转载)
本文结构 ASP.NET Core应用程序的构建 ASP.NET Core应用程序容器化所需注意的问题 应用程序的配置信息 端口侦听 ASP.NET Core的容器版本 docker镜像构建上下文(B ...
- CentOS7 firewalld防火墙 启动 关闭 禁用 添加删除规则等 常用命令
CentOS7 firewalld防火墙 常用命令1.firewalld的基本使用启动: systemctl start firewalld关闭: systemctl stop firewalld查看 ...
- 对ssm框架里面的一些常用注解的理解
@Componcnt :作用就是把当前类对象存入spring容器中 属性:value 用于指定bean的id 当我们不写的时候默认就是当前类名,并且首字母要小写 ------------------- ...
- Debian忘记密码重置
前一阵子因为特殊原因我把一个网站的VPS服务器关闭了,结果竟把SSH密码忘了,也没有使用SSH密钥,因为上面还有网站文件不能选择重装,只能尝试在面板重置,但是面板返回结果一直是404我无法获得重置的密 ...
- Windows下载安装RabbitMQ教程
原文链接:http://www.studyshare.cn/software/details/1171/0一.下载 1.下载Erlang 官网下载:去下载 百度网盘下载:去下载 提取码:m1q0 2 ...
- 【JVM】CMS垃圾回收器
一.简介 Concurrent Mark Sweep,是一种以获取最短回收停顿时间为目标的收集器,尤其重视服务的响应速度. CMS是老年代垃圾回收器,基于标记-清除算法实现.新生代默认使用ParNew ...