Problem:
Score : 400 points

Problem Statement

For a positive integer X, let f(X) be the number of positive divisors of X.

Given a positive integer N, find \sum_{K=1}^N K\times f(K).

Constraints

1 \leq N \leq 10^7

Input

Input is given from Standard Input in the following format:

N

Output

Print the value \sum_{K=1}^N K\times f(K).

Sample Input 1

4

Sample Output 1

23

We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\times 1 + 2\times 2 + 3\times 2 + 4\times 3 =23.

Sample Input 2

100

Sample Output 2

26879

Sample Input 3

10000000

Sample Output 3

838627288460105

Watch out for overflows.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<limits.h>
#include<assert.h>
#include <unordered_set>
#include <unordered_map>

#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)
#define FORS(I, S) for (int I = 0; S[I]; ++I)
#define RS(X) scanf("%s", (X))
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PII> VPII;
typedef pair<LL,LL> PLL;
typedef vector<PLL> VPLL;
typedef set<LL> SLL;
typedef unordered_set<LL> USLL;
typedef string STR;

template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(LL &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }
template<class T> void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const LL &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template<class T,class U> void _W(const pair<T,U> &x) {_W(x.F); putchar(' '); _W(x.S);}
template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }
void W() {}
template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }
#ifdef HOME
 #define DEBUG(...) {printf("# ");printf(__VA_ARGS__);puts("");}
#else
 #define DEBUG(...)
#endif
int MOD = 1e9+7;

std::pair<LL, LL> find_rest(LL n, LL dev) {
  LL degree = 0;
  while (n % dev == 0) {
    n /= dev;
    ++degree;
  }
  return {n, degree};
}

int main() {
  LL n;
  R(n);
  std::vector<LL> devs(n + 1, -1);
  devs[1] = 1;
  for (LL i = 2; i <= n; ++i) {
    if (devs[i] != -1) {
      continue;
    }

    for (LL j = i; j <= n; j += i) {
      devs[j] = i;
    }
  }

  std::vector<LL> f(n + 1);

  f[1] = 1;
  for (LL i = 2; i <= n; ++i) {
    if (devs[i] == i) {
      f[i] = 2;
    } else {
      auto [rest, degree] = find_rest(i, devs[i]);

      f[i] = f[rest] * (degree + 1);
    }
  }

  LL res = 0;

  for (LL i = 1; i <= n; ++i) {
    res += i * f[i];
  }

  W(res);

  return 0;
}