1. Abridged problem statement
- There are n participants and m football matches.
- For each match, the actual result (a, b) is given, followed by n predictions (p_i, q_i), one per participant.
- Scoring per prediction:
  - +2 if the outcome (win/draw/loss) is correct.
  - +3 if the goal difference a - b is correct.
  - +1 if the first team's goals a are correct.
  - +1 if the second team's goals b are correct.
- Output the total score for each participant after all matches.

2. Detailed editorial
- Observations:
  - The score is additive across four independent checks per match: outcome, difference, first team goals, second team goals.
  - If the exact score is guessed, all four conditions hold, awarding 2 + 3 + 1 + 1 = 7.
  - If the difference is correct, the outcome is automatically correct as well, including the draw case (difference 0). Both bonuses are still awarded, per the rules.
- Algorithm:
  - Read n, m.
  - For each match:
    - Read actual result (a, b).
    - For each participant i:
      - Read prediction (p, q).
      - Initialize score = 0.
      - Outcome: if (a > b and p > q) or (a < b and p < q) or (a == b and p == q), add 2.
      - Difference: if (a - b) == (p - q), add 3.
      - First team goals: if a == p, add 1.
      - Second team goals: if b == q, add 1.
      - Accumulate to participant i's total.
  - Print all totals.
- Correctness:
  - The above directly implements the scoring rules independently. Every condition is checked exactly as specified.
- Complexity:
  - Time: O(n*m), at most 100*100 = 10,000 predictions; trivial within limits.
  - Memory: O(n*m) if all predictions are stored; can also be done streaming with O(n) memory, but not necessary.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m;
vector<pair<int, int>> actual;
vector<vector<pair<int, int>>> predictions;

void read() {
    cin >> n >> m;
    actual.resize(m);
    predictions.resize(m);

    for(int match = 0; match < m; match++) {
        cin >> actual[match];
        predictions[match].resize(n);
        cin >> predictions[match];
    }
}

void solve() {
    // The solution here is to directly implement what's described in the
    // problem. There is no real trick or insight but rather following the rules
    // in the problem statement.

    vector<int> scores(n, 0);

    for(int match = 0; match < m; match++) {
        int a = actual[match].first;
        int b = actual[match].second;

        for(int i = 0; i < n; i++) {
            int p = predictions[match][i].first;
            int q = predictions[match][i].second;

            int score = 0;

            // Check if guessed the winner (or tie)
            if((a > b && p > q) || (a < b && p < q) || (a == b && p == q)) {
                score += 2;
            }

            // Check if guessed the difference
            if(a - b == p - q) {
                score += 3;
            }

            // Check if guessed first team's goals
            if(a == p) {
                score += 1;
            }

            // Check if guessed second team's goals
            if(b == q) {
                score += 1;
            }

            scores[i] += score;
        }
    }

    cout << scores << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```python
import sys

def main():
    data = list(map(int, sys.stdin.read().split()))  # Read all integers from stdin.
    it = iter(data)                                   # Create an iterator for sequential reading.

    try:
        n = next(it)                                  # Number of participants.
        m = next(it)                                  # Number of matches.
    except StopIteration:
        return                                        # Empty input; nothing to do.

    # Initialize total scores for all participants.
    scores = [0] * n

    # Process each match.
    for _ in range(m):
        # Read the actual result (a, b).
        a = next(it)
        b = next(it)

        # For each participant, read prediction and compute score contribution.
        for i in range(n):
            p = next(it)
            q = next(it)

            add = 0

            # Outcome correct (winner or draw)?
            if (a > b and p > q) or (a < b and p < q) or (a == b and p == q):
                add += 2

            # Difference correct?
            if (a - b) == (p - q):
                add += 3

            # First team's goals correct?
            if a == p:
                add += 1

            # Second team's goals correct?
            if b == q:
                add += 1

            scores[i] += add

    # Output the scores separated by spaces.
    print(' '.join(map(str, scores)))

if __name__ == "__main__":
    main()
```

5. Compressed editorial
- For each match and participant, compute four independent bonuses:
  - +2 if outcome (win/draw/loss) matches,
  - +3 if goal difference matches,
  - +1 if first team goals match,
  - +1 if second team goals match.
- Sum these across all matches per participant and print. Complexity O(n*m), with trivial constants.
