Sum of MSLCM

Правка en2, от BumbleBee, 2018-02-09 18:51:02

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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en2 Английский BumbleBee 2018-02-09 18:51:02 6
en1 Английский BumbleBee 2018-02-09 18:06:54 1548 Initial revision (published)