### The_mysterio's blog

By The_mysterio, history, 7 weeks ago,

Hello all, I am trying to find the complexity of the below code

"#include <bits/stdc++.h>

using namespace std;

int main() {

long long int n;

cin>>n;

vector<int>factors[1000001];

for(int i=2;i<=n;i++)

{

for(int j=1;j*i<=n;j++)

{

factors[j*i].push_back(i);

}

}

return 0;


}"

What I am mainly doing in the code is trying to store factors of all the numbers from 1 to n (n<=1000000). As you can see , factors is a vector of vectors. factors[i] will store all the factors for number i. The code works as follows 1. Fix a number i in the outer loop 2. For all the multiples of i that lies within n, push i as it's factor-->this is being done in the inner loop. According to my calculation, the complexity should be nlog(n) which should easily pass in a time limit of 1s but when I am running the above code on Codechef compiler with n=1000000, the execution time is coming >2s.

If anyone can kindly help, I will be highly obliged. Thank you..

• -12

 » 7 weeks ago, # |   +5 it could be because vector push_back is too slow, especially since the size of the each vector is not very large (average size is about 14 factors)
•  » » 7 weeks ago, # ^ |   0 Thank you for your concern.So what should be the remedy? Use vector.reserve?
•  » » » 7 weeks ago, # ^ |   0 i suppose that it would speed it up, not sure which number is most optimal i tested 15 and 30, both of them gave ~850ms on cf (while without reserve gave ~1000ms)
•  » » » » 7 weeks ago, # ^ |   0 With the following snippet there are no reallocations of vectors:  for (size_t i = 2; i <= n; ++i) factors[i].reserve(sqrt(i)+5); 
•  » » » » » 7 weeks ago, # ^ |   0 it also MLEs? O(n sqrt(n)) memory
 » 7 weeks ago, # |   0 It is also not cache-friendly. Storing multiples is much faster than storing divisors.
 » 7 weeks ago, # | ← Rev. 3 →   0 AFAIU, if x is a divisor of n then all divisors of x are also divisors of n. So it is not needed to duplicate work as well as to store divisors multiple times, is it?Maybe this code could help: Spoiler for (unsigned i = 2; i <= n; ++i) { for (unsigned j = 2, x = i; j*j <= x; ++j) { if (i % j == 0) { if (j*j != i) factors[i].push_back(j); x = i / j; factors[i].push_back(x); } } factors[i].push_back(i); } 
 » 7 weeks ago, # |   0 Thank you everyone for your concern. Well, I used the code by user "actium" giving me a runtime of 1.44 s. Guess I have to solve the problem in other way. Thank you everyone.