codeforces C. Everyone is a Winner!

题目要求参赛者在codeforces的评级分配问题中找出所有可能的评级增加值。给定总评级单位n,需要找到所有可能通过整除n得到的评级增量,并按升序输出。解题思路包括使用set和C++11的auto特性来避免内存超限和运行超时。

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原题链接:codeforces C. Everyone is a Winner!

题目要求

On the well-known testing system MathForces, a draw of nn rating units is arranged. The rating will be distributed according to the following algorithm: if kk participants take part in this event, then the nn 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=5n=5 and k=3k=3, then each participant will recieve an 11 rating unit, and also 22 rating units will remain unused. If n=5n=5, and k=6k=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=5n=5, then the answer is equal to the sequence 0,1,2,50,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋⌊n/k⌋ for some positive integer kk (where ⌊x⌋⌊x⌋ is the value of xx rounded down): 0=⌊5/7⌋0=⌊5/7⌋, 1=⌊5/5⌋1=⌊5/5⌋, 2=⌊5/2⌋2=⌊5/2⌋, 5=⌊5/1⌋5=⌊5/1⌋.

Write a program that, for a given nn, finds a sequence of all possible rating increments.

Input
The first line contains integer number tt (1≤t≤101≤t≤10) — the number of test cases in the input. Then tt test cases follow.

Each line contains an integer nn (1≤n≤1091≤n≤109) — the total number of the rating units being drawn.

Output
Output the answers for each of tt test cases. Each answer should be contained in two lines.

In the first line print a single integer mm — the number of different rating increment values that Vasya can get.

In the following line print mm integers in ascending order — the values of possible rating increments.

Example
input
4
5
11
1
3
output
4
0 1 2 5
6
0 1 2 3 5 11
2
0 1
3
0 1 3

解题思路

这题是真不难,但坑就坑在总是超内存 大哭
给大家看一下我的战绩:
在这里插入图片描述
Memory Limit Exceeded(内存超限)
Runtime Error(运行超时)
太坑爹了,我之前的代码是这样的:可以过前三块样例

#include <iostream>
### Codeforces Div.2 比赛难度介绍 Codeforces Div.2 比赛主要面向的是具有基础编程技能到中级水平的选手。这类比赛通常吸引了大量来自全球不同背景的参赛者,包括大学生、高中生以及一些专业人士。 #### 参加资格 为了参加 Div.2 比赛,选手的评级应不超过 2099 分[^1]。这意味着该级别的竞赛适合那些已经掌握了一定算法知识并能熟练运用至少一种编程语言的人群参与挑战。 #### 题目设置 每场 Div.2 比赛一般会提供五至七道题目,在某些特殊情况下可能会更多或更少。这些题目按照预计解决难度递增排列: - **简单题(A, B 类型)**: 主要测试基本的数据结构操作和常见算法的应用能力;例如数组处理、字符串匹配等。 - **中等偏难题(C, D 类型)**: 开始涉及较为复杂的逻辑推理能力和特定领域内的高级技巧;比如图论中的最短路径计算或是动态规划入门应用实例。 - **高难度题(E及以上类型)**: 对于这些问题,则更加侧重考察深入理解复杂概念的能力,并能够灵活组合多种方法来解决问题;这往往需要较强的创造力与丰富的实践经验支持。 对于新手来说,建议先专注于理解和练习前几类较容易的问题,随着经验积累和技术提升再逐步尝试更高层次的任务。 ```cpp // 示例代码展示如何判断一个数是否为偶数 #include <iostream> using namespace std; bool is_even(int num){ return num % 2 == 0; } int main(){ int number = 4; // 测试数据 if(is_even(number)){ cout << "The given number is even."; }else{ cout << "The given number is odd."; } } ```
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