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）