Approach for problem

[big_b1]

Could anyone tell me the approach to solve the following math problem in CSES CSES - Sum of Divisors

[everule1]

Sum of divisors of all numbers less than n in less than O(n) time

It’s just \sum_{i=1}^n i\lfloor n/i\rfloor Since i divides \lfloor n/i \rfloor values less than n. To optimise this we can notice that \lfloor n/i\rfloor goes through only O(\sqrt{n}) values. n/j is i for j\in (\lfloor n/(i+1) \rfloor, \lfloor n/i \rfloor \ ]

My code

#include <iostream> #include <bits/stdc++.h> #include <cmath> #include <vector> #define ll long long int #define mp make_pair #define pb push_back #define vi vector<int> using namespace std; const ll…

[ssrivastava990]

Superb problem

[ssrivastava990]

Here is mine : Solution

[big_b1]

Thanks !! Got it