By sadman.rizwan, history, 6 months ago, ,

Recently I was trying to solve the problem Sum of MSLCM from ACM ICPC Live Archive.

After a little analysis, I found out that the answer is S — 1, where S is the sum of all X such that,

X = A * (int)(N / A) for all A in the range [1, N].

int ans=0
for(int a=1;a<=n;a++){
ans+=a*(int)(n/a);
}


But my solution won't pass the time limit if I iterate over the range [1, N] as N can be as large as 20000000.

I couldn't find another solution even after several hours of thinking and decided to search for the solution on the internet. While searching for the solution I found the following code.

#include <bits/stdc++.h>
using namespace std;
#define long long long int
int main()
{
ios_base::sync_with_stdio(false);
long n;
while(cin>> n && n){
long ans=0,a=1;
while(a<=n){
long x=n/a;
long y=n/x;
ans+=x*(((a+y)*(y-a+1))/2);
a=++y;
}
ans--;
cout<< ans <<endl;
}
return 0;
}


This solution passed the time limit for the given input range and got accepted.

But I still couldn't understand the formula used in this code. The only optimization I see here is that it doesn't iterate over all the numbers in the range [1, N]. But how does that work?

I want a mathematical explanation of this solution.

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• +8
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 » 6 months ago, # | ← Rev. 3 →   +9 Let , 1 ≤ x ≤ n. Let's find out how many distinct values the function can get by dividing it in two cases. Case 1: If , f can get at most distinct values. Case 2: If , n / x is always less than or equal to . Therefore, f can get at most distinct values. We can find the values and count their appearances in time, which allows us to compute the sum efficiently.
•  » » 6 months ago, # ^ |   0 Well understood. thanks a lot.
 » 6 months ago, # | ← Rev. 2 →   0 I know this is not the expected solution, but it can be solved in O(n·log n) time and O(n) memory. Doing a preprocessing with n = 2·107 and with O(1) queries the execution time is almost five seconds on the server, but it still passes.
•  » » 6 months ago, # ^ |   +3 I tried preprocessing to answer each query in O(1) time, but couldn't. Can you explain in detail how it can be done?
•  » » » 6 months ago, # ^ |   0 You can apply a sieve, suppose S[i] =  Sum of Divisors of i.For every number i in [1..107] we add i to its multiples (2i, 3i, 4i,...)Then, we apply a preffix sum (S[i] = S[i] + S[i - 1]).This preprocessing has a complexity of O(n·log n) where n is the maximum number in the tests. The constraints state that 2 ≤ n ≤ 107Again, this is not the expected solution, and I think it's not even a good one!
•  » » » » 6 months ago, # ^ |   0 Got this. It's not the fastest solution, but it may pass the time limit I think.