Ah, yes, competitive programming... This would be rated codeforces-800

Challenge files: mvm.pdf

Continuing with last year's PPC success, I tackled this first. Too bad ours is only the 4th fastest solve...

The challenge

We need to select a fixed show time for m shows and divide them equally among n artists, in a way such that no show have more than one artist transition.

n <= m

This should have some eyebrows raised. Coupled with the fact that: if an artist's total show time is less than an individual show's time, there will be a show where there are two artist transitions; and we have our idea: select n as the length of each show, and have each artist's show time be m. Afterwards we can manually assign the artist(s) for each show in O(n) time.

But the "fact" you stated is...

For a longer post, a visual proof of this will be shown below.

pardon my incredible art skills

The C++ implementation

I find the code self-explanatory so...

#include <bits/stdc++.h>

using namespace std;

int n, m;

void solve() {
    int currentArtist = 1, remainingArtistTime = m;
    cout << n << "\n";
    for (int i = 1; i <= m; i++) {
        if (n <= remainingArtistTime) {
            cout << n << " " << currentArtist << " 0 " << currentArtist << "\n";
            remainingArtistTime -= n;
        } else {
            cout << remainingArtistTime << " " << currentArtist << " "
            << n - remainingArtistTime << " " << currentArtist + 1 << "\n";
            currentArtist++;
            remainingArtistTime = m - (n - remainingArtistTime);
        }
    }
}

int main() {
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        cin >> n >> m;
        solve();
    }
    return 0;
}

Now excuse me, I need to sleep, the time is 11:15 pm...