sadman.rizwan's blog

By sadman.rizwan, history, 8 months ago, In English,

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 months ago, # |
Rev. 3   Vote: I like it +9 Vote: I do not like it

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.

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8 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

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.

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    8 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I tried preprocessing to answer each query in O(1) time, but couldn't. Can you explain in detail how it can be done?

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      8 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      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 ≤ 107

      Again, this is not the expected solution, and I think it's not even a good one!

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        8 months ago, # ^ |
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        Got this. It's not the fastest solution, but it may pass the time limit I think.