Codeforces1062D. Fun with Integers（埃氏筛）

题目链接：传送门

题目：

D. Fun with Integers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a positive integer n
greater or equal to 2. For every pair of integers a and b (2≤|a|,|b|≤n), you can transform a into b if and only if there exists an integer x such that 1<|x| and (a⋅x=b or b⋅x=a), where |x| denotes the absolute value of x

.

After such a transformation, your score increases by |x|
points and you are not allowed to transform a into b nor b into a

anymore.

Initially, you have a score of 0

. You can start at any integer and transform it as many times as you like. What is the maximum score you can achieve?
Input

A single line contains a single integer n
(2≤n≤100000

) — the given integer described above.
Output

Print an only integer — the maximum score that can be achieved with the transformations. If it is not possible to perform even a single transformation for all possible starting integers, print 0

.
Examples
Input
Copy

4

Output
Copy

8

Input
Copy

6

Output
Copy

28

Input
Copy

2

Output
Copy

0

Note

In the first example, the transformations are 2→4→(−2)→(−4)→2

.

In the third example, it is impossible to perform even a single transformation.

思路：

　　如果一对数a，b可以通过x转换得到分数，那么a→b→-a→-b→a总共可以得到4*|x|的分数。

　　不妨只考虑0 < a < b，0 < x的情况，把结果乘上4就好了。

　　此时只能有a*x = b，即b为a的整数倍。对于给定的a，x可以取[2, n/a]范围内的所有值，为一个等差数列。用求和公式计算即可。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main()
{
    ll n;
    cin >> n;
    ll ans = 0;
    for (ll i = 2; i <= n/2; i++) {
        ll cnt = n/i;
        ans += 4*(2+cnt)*(cnt-1)/2;
    }
    cout << ans << endl;
    return 0;
}