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admin 11月 14, 2020 0

D. Fun with Integers

题意

You are given a positive integer n greater or equal to 2. For every pair of integers a and b (2≤|𝑎|,|𝑏|≤𝑛2≤|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?

题解

答案为 2~n 每个数的因子( 范围为$[2,n)$ )之和。

$Hugesum s_i$

代码


#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define inlld(lld) scanf("%lld", &lld)
#define inlf(f) scanf("%lf", &f)
#define ind(d) scanf("%d", &d)
#define ins(s) scanf("%s", s)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef std::pair<long long, long long> pll;
typedef std::pair<int, int> pii;
typedef unsigned long long ull;
typedef long double db;
typedef long long ll;
const db pi = acos(-1.0);
const ll inf =0x3f3f3f3f;
const ll mod = 1e9+7;
const int N = 1.1e5;
const db eps = 1e-8;
using namespace std;
int (db a) { return a < -eps ? -1 : a > eps; }
int db_cmp(db a, db b){ return sign(a-b); }

int main() {
#ifndef ONLINE_JUDGE
    
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n;
    ll ans=0;
    cin>>n;
    forl(i,2,n){
        ans+=1ll*i*(n/i-1);
    }
    cout<<ans*4<<endl;
    return 0;
}