According to the problem statement of problem 1091G - New Year and the Factorisation Collaboration, "the run time of the interactor is also counted towards the time limit." So, I'm pretty sure that the TL includes the time taken by the interactor.

Both the interactor and the participant's solution run simultaneously and none of them can proceed without the other's response. So, the execution time for both of them must be equal to the total time of the complete interaction. In your case, any participant's solution will have to wait (stay idle) for the 500ms that the interactor will take and this will probably be counted towards the execution time of the participant's solution.
SPOJ takes care of these issues by assigning the actual time limit as 2 times the given time limit plus some constant for both participant's solution and interactor. You can read about it here. I am not sure if Polygon does the same thing or not but if I have to guess then it probably doesn't as the intended solution for one of the interactive problem which I had set on Codechef took around 1500ms on Polygon and around 400ms on Codechef (which uses SPOJ libraries).

I've checked that in Polygon.

Problem statement: given a number $$$n$$$, you need to print $$$\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^n i \cdot j \right) \mod (10^9+7)$$$. You can ask query ? any number of times, then each time interactor will calculate the answer and give it to solution. At the end you need to print ! ans.

#include "testlib.h"
#include <iostream>

using namespace std;

const int MOD=1e9+7;

int calc(int n)
{
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            ans=(ans+1ll*i*j)%MOD;
    return ans;
}

int main(int argc, char* argv[])
{
    registerInteraction(argc, argv);
    
    int n = inf.readInt();
    cout<<n<<endl;

    while(true)
    {
        string s=ouf.readWord();
        if(s=="!")
        {
            int ans=ouf.readInt();
            tout<<ans;
            quitf(_ok,"%d",ans);
        }
        else if(s=="?")
            cout<<calc(n)<<endl;
        else
            quitf(_wa,"unknown query \"%s\"",s.c_str());
    }
}

I have few solutions, they start with the following code:

#include<bits/stdc++.h>

using namespace std;

const int MOD=1e9+7;

int calc(int n)
{
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            ans=(ans+1ll*i*j)%MOD;
    return ans;
}

void ask()
{
    cout<<"?"<<endl;
}

int read()
{
    int ans;
    cin>>ans;
    return ans;
}

They differ with main function:

    int n;
    cin>>n;
    
    ask();
    int r=read();
    
    cout<<"! "<<r<<endl;

    int n;
    cin>>n;
    
    int c=calc(n);
    
    cout<<"! "<<c<<endl;

    int n;
    cin>>n;
    
    int c=calc(n);
    ask();
    int r=read();
    
    cout<<"! "<<(c+r)/2<<endl;

    int n;
    cin>>n;
    
    ask();
    int c=calc(n);
    int r=read();
    
    cout<<"! "<<(c+r)/2<<endl;

    int n;
    cin>>n;
    
    ask();
    int r=read();
    int c=calc(n);
    
    cout<<"! "<<(c+r)/2<<endl;

Problem has $$$14$$$ tests, $$$i$$$-th of them is $$$n=5000 \cdot i$$$. TL is 5 s.

Invocation results

My conclusions (I am not sure, but it looks like this):

• displayed time is just solution's runtime;

• if just solution's runtime is bigger than TL then you get TL;

• if just interactor's runtime is bigger than some value then you get IL;

• some value is a bit (2 times?) bigger than TL;

• sum of just solution's runtime and just interactor's runtime isn't calculated.