Torni

input/code.cpp: In function 'std::vector<long long int> test(ll, std::vector<long long int>)':
input/code.cpp:12:22: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
    for (int i = 1; i < as.size(); ++i) {
                    ~~^~~~~~~~~~~
input/code.cpp:19:17: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
   if (as.size() > n) return {};
       ~~~~~~~~~~^~~
input/code.cpp:23:16: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
  if (as.size() < n) return {};
      ~~~~~~~~~~^~~
input/code.cpp: In function 'void solve()':
input/code.cpp:65:4: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
    for (auto v : res) cout << v + minv << ' '; cout << '\n';
    ^~~
input/code.cpp:65:48: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
    for (auto v : res) cout << v + minv << ' '; cout << '\n';...

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = (ll)1e9 + 7;

vector<ll> test(ll n, vector<ll> bs) {
	vector<ll> as = {0};
	multiset<ll> expected;
	for (ll v : bs) {
		auto it = expected.find(v);
		if (it == expected.end()) {
			for (int i = 1; i < as.size(); ++i) {
				expected.insert(as[i] + v);
			}
			as.push_back(v);
		} else {
			expected.erase(it);
		}
		if (as.size() > n) return {};
	}

	// Check validity
	if (as.size() < n) return {};
	for (int i = 0; i+1 < n; ++i) {
		if (as[i] > as[i+1]) return {};
	}

	vector<ll> sums;
	for (int i = 0; i < n; ++i) {
		for (int j = i+1; j < n; ++j) {
			sums.push_back(as[i] + as[j]);
		}
	}
	sort(sums.begin(), sums.end());
	if (sums != bs) return {};

	return as;
}

void solve() {
	int n;
	cin >> n;
	int m = n*(n-1) / 2;

	vector<ll> bs(m);
	for (ll& v : bs) cin >> v;
	sort(bs.begin(), bs.end());

	// If we were guaranteed that the minimum would be zero, the problem would be easy to solve
	// Can we find the minimum value in some n^2 attempts? YES:
	//	Smallest value must equal bs[0] = as[0] + as[1]
	//	Second smallest must equal bs[1] = as[0] + as[2]
	//	Loop which value equals bs[2] = as[1] + as[2].
	//		-> At worst, the nth pair!
	//	Minimum is 2*bs[0] - bs[1] - bs[2]
	//	Subtract twice that from all values

	for (int i = 2; i < n; ++i) {
		ll minv = (bs[0] + bs[1] - bs[i]) / 2;
		vector<ll> tmp = bs;
		for (ll& v : tmp) v -= 2*minv;

		auto res = test(n, tmp);
		if (! res.empty()) {
			for (auto v : res) cout << v + minv << ' '; cout << '\n';
			return;
		}
	}
	cout << "-1\n";
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	solve();
}

Test 1

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 2

Group: 2, 3

Verdict: WRONG ANSWER

Test 3

Group: 3

Verdict: WRONG ANSWER