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
- FPGA控制RGMII接口PHY芯片基础
一.前言 网络通信中的PHY芯片接口种类有很多,之前接触过GMII接口的PHY芯片RTL8211EG.但GMII接口数量较多,本文使用RGMII接口的88E1512搭建网络通信系统.这类接口总线位宽小 ...
- golang ---rune与byte
golang内置类型有rune类型和byte类型. rune类型的底层类型是int32类型,而byte类型的底层类型是int8类型,这决定了rune能比byte表达更多的数. 在unicode中,一个 ...
- PAT 1003我要通过!
PAT 1003 我要通过! 答案正确"是自动判题系统给出的最令人欢喜的回复.本题属于 PAT 的"答案正确"大派送 -- 只要读入的字符串满足下列条件,系统就输出&qu ...
- Devops K8s
公司在组建Devops团队,base在上海 徐家汇.具体职位有Devops工程师和K8s工程师. 有意者请私信.
- selenium 开启开发者工具(F12)
selenium 开启开发者工具(F12) options = webdriver.ChromeOptions(); options.add_argument("--auto-open-de ...
- Java 流程控制语句 之 顺序结构
在一个程序执行的过程中,各条语句的执行顺序对程序的结果是有直接影响的.也就是说,程序的流程对运行结果 有直接的影响.所以,我们必须清楚每条语句的执行流程.而且,很多时候我们要通过控制语句的执行顺序来实 ...
- MongoDB用户和密码登录
一.MongoDB中内置角色 角色 介绍 read 提供读取所有非系统的集合(数据库) readWrite 提供读写所有非系统的集合(数据库)和读取所有角色的所有权限 dbAdmin 提供执行管理任务 ...
- WinForm背景图片及图片位置
设置背景图片:BackgroundImage属性选择对应的图片就可以了. 背景图片随窗体的变化而变化:BackgroundImageLayout属性值设置为Stretch. 窗体放置图片:Pictur ...
- Django之form主键
Form介绍 我们之前在HTML页面中利用form表单向后端提交数据时,都会写一些获取用户输入的标签并且用form标签把它们包起来. 与此同时我们在好多场景下都需要对用户的输入做校验,比如校验用户是否 ...
- Odoo字段类型详解
转载请注明原文地址:https://www.cnblogs.com/ygj0930/p/10826099.html 一:基本字段类型 Binary:二进制类型,用于保存图片.视频.文件.附件等,在 ...