Good Bye 2018 C. New Year and the Sphere Transmission(數學)

題目鏈接：C. New Year and the Sphere Transmission

There are $n$ people sitting in a circle, numbered from $1$ to $n$ in the order in which they are seated. That is, for all $i$ from $1$ to $n-1$, the people with id $i$ and $i+1$ are adjacent. People with id $n$ and $1$ are adjacent as well.

The person with id $1$ initially has a ball. He picks a positive integer $k$ at most $n$, and passes the ball to his $k$-th neighbour in the direction of increasing ids, that person passes the ball to his $k$-th neighbour in the same direction, and so on until the person with the id $1$ gets the ball back. When he gets it back, people do not pass the ball any more.

For instance, if $n = 6$ and $k = 4$, the ball is passed in order $[1, 5, 3, 1]$.

Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be $1 + 5 + 3 = 9$.

Find and report the set of possible fun values for all choices of positive integer $k$. It can be shown that under the constraints of the problem, the ball always gets back to the $1$-st player after finitely many steps, and there are no more than $10^5$ possible fun values for given $n$.

Input

The only line consists of a single integer $n$ ($2 \leq n \leq 10^9$) — the number of people playing with the ball.

Output

Suppose the set of all fun values is $f_1, f_2, \dots, f_m$.

Output a single line containing $m$ space separated integers $f_1$ through $f_m$ in increasing order.

Examples

Input

6

Output

1 5 9 21

Input

16

Output

1 10 28 64 136

Note

In the first sample, we’ve already shown that picking $k = 4$ yields fun value $9$, as does $k = 2$. Picking $k = 6$ results in fun value of $1$. For $k = 3$ we get fun value $5$ and with $k = 1$ or $k = 5$ we get $21$.

In the second sample, the values $1$, $10$, $28$, $64$ and $136$ are achieved for instance for $k = 16$, $8$, $4$, $10$ and $11$, respectively.

思路

有$n$個數圍成一個環，編號是從$1$到$n$ ，現在你可以隨意選擇一個不超過$n$的數$k$，從$1$開始，按照環上字符增大的順序，每隔$k$個選擇一個數字，直到回到起點$1$，然後計算出路徑上不同數字的和，記爲fun value，題目給出你一個數$n$，然後讓你求出這$n$個數的所有的fun value，從小到大輸出結果。

如果選擇的$k$是$n$的因數，那麼一定可以一輪就走完，不同的因數，獲得的fun value的值也不一樣。

當選擇的$k$不是$n$的因數時，那麼一輪一定走不完，然後直到某一輪走的路徑正好和$k$是$n$的因數的某一條路徑重合，所以我們計算$k$是$n$的因數的路徑時，已經把這種情況包含了。

所以，利用等差數列求和公式$S_n=\frac{n(a_1+a_n)}{2}$可以計算出結果。第一項一定爲$1$，數字的個數就是當前枚舉的$i$或者$\frac{n}{i}$.

可以打個表驗證一下:

打表代碼

#include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
int get_x(int n, int x)
{
    if (x == n)
        return 1;
    int s = x, sum = x + 1;
    while (s != 0)
    {
        s = (s + x) % n;
        sum += (s + 1);
    }
    return sum;
}
set<int> getall_n(int n)
{
    set<int> s;
    for (int i = 1; i <= n; i++)
        s.insert(get_x(n, i));
    return s;
}
int main()
{
    for (int i = 1; i <= 100; i++)
    {
        printf("n=%d:\n", i);
        set<int> s = getall_n(i);
        for (auto x : s)
            printf("%d ", x);
        printf("\n");
    }
    return 0;
}

然後印證了我們的觀點，每個數的答案個數就是他的因子個數。

代碼

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
set<ll> s;
int main()
{
    ll n;
    scanf("%lld", &n);
    for (ll i = 1; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            ll x = n / i;
            s.insert((1 + n - i + 1) * x / 2);
            s.insert((1 + n - x + 1) * i / 2);
        }
    }
    for (auto x : s)
        printf("%lld ", x);
    puts("");
    return 0;
}