Consider the list $x$ for a single test case.

Solution 1: Solve $\frac{7!}{(7-N)!}$ distinct linear systems using Gaussian elimination. But of course, Silver contestants are not expected to know how to do this.

Solution 2: Add $0$ to $x$, so $x$ now contains between $5$ and $8$ elements inclusive. Suppose that $x$ is compatible with a triple $(A,B,C)$. Then consider the following pairs of integers:

• $(0,A)$
• $(B,A+B)$
• $(C,A+C)$
• $(B+C,A+B+C)$

Since the size of $x$ is greater than $4$, $x$ must contain two elements from the same pair, implying that there exist two elements of $x$ such that their difference equals $A$. Similar reasoning holds for $B$ and $C$.

So it suffices to iterate through all possibilities for $A$, $B$, and $C$ and increment the answer whenever we come across a candidate triple that works.

Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class DoYouKnowYourABCsCounting {
 
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder out = new StringBuilder();
        int t = Integer.parseInt(in.readLine());
        for (int tc = 1; tc <= t; tc++) {
            int n = Integer.parseInt(in.readLine());
            int[] numbers = new int[n];
            StringTokenizer tokenizer = new StringTokenizer(in.readLine());
            for (int j = 0; j < n; j++) {
                numbers[j] = Integer.parseInt(tokenizer.nextToken());
            }
            int answer = 0;
            Set<Integer> expanded = new HashSet<>();
            for (int x : numbers) {
                expanded.add(x);
                for (int y : numbers) {
                    if (x < y) {
                        expanded.add(y - x);
                    }
                }
            }
 
            for (int a : expanded) {
                for (int b : expanded) {
                    for (int c : expanded) {
                        if (a <= b && b <= c) {
                            List<Integer> allNumbers = Arrays.asList(a, b, c, a + b, b + c, c + a, a + b + c);
                            boolean works = true;
                            for (int x : numbers) {
                                if (!allNumbers.contains(x)) {
                                    works = false;
                                }
                            }
                            if (works) {
                                answer++;
                            }
                        }
                    }
                }
            }
            out.append(answer).append('\n');
        }
        System.out.print(out);
    }
}

Solution 3: There exist two elements of $x$ such that their sum equals $A+B+C$ (by similar reasoning as solution 2). Given the sum, we only need to check two candidate solutions.

My code:

#include <bits/stdc++.h>
using namespace std;

int N;
vector<int> x;
set<multiset<int>> sols;

void check_sol(int sum, int b, int c) {
	int a = sum-b-c;
	assert(a > 0 && b > 0 && c > 0);
	set<int> s{0,a,b,c,a+b,b+c,c+a,a+b+c};
	for (int t: x) if (!s.count(t)) return;
	sols.insert({a,b,c});
}
 
void test(int sum) {
	set<int> candidates;
	for (int t: x) {
		if (t > sum) return;
		if (t == 0 || t == sum) continue;
		candidates.insert(min(t,sum-t));
	}
	assert(candidates.size() >= 2);
	int a = *begin(candidates);
	int b = *next(begin(candidates));
	check_sol(sum,a,b);
	check_sol(sum,a,sum-b);
}
 
int solve() {
	cin >> N; 
	x.resize(N); for (int& t : x) cin >> t;
	x.push_back(0);

	sols.clear();
	for (int i = 0; i < (int)x.size(); ++i)
		for (int j = i+1; j < (int)x.size(); ++j)
			test(x[i]+x[j]);
	return (int)sols.size();
}
 
int main() {
	int T; cin >> T;
	while (T--) cout << solve() << "\n";
}