Known set pairwise sum

topic

Given the sum of the two elements in the set, find the elements in the set

Ideas

Draw a matrix first

$a_1$ + $a_2$	$a_1$ + $a_3$	$a_1$ + $a_4$
-	$a_2$ + $a_3$	$a_2$ + $a_4$
-	-	$a_3$ + $a_4$

Among them $a_1,a_2...$ It is in ascending order of
easy to know, by the single matrix value from left to right, top to bottom monocytogenes.
So if we can get the numbersa 1, a 2,..., An a_1, a_2,..., a_nthat appeared in the first n columns first$a_1,a_2,...,a_n$
A first $n+The$ first element in column$1$ is the smallest element in the remaining sum,$a_{n+1}$It was confirmed.
With the same induction method, all results can be obtained (maybe it can be called recursion?)

Then only need to traverse $a_1,a_2$Possible value range, and then inductively check whether the remaining sum can satisfy $a_3,...,a_n$If you can, it is a set of answers

The complexity should be $O(S_1.(F_S{)}^N)$
where$S_1$Is the minimum sum value, $N$ is the number of set elements,$F_S$Is the complexity required to find the minimum value, the set I used, here is $LogN$

Attach code

#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
#define CNT 5051
int N;
multiset<int> sum;
int ans[101];
bool flag = 0;
void SolveLayer(int layer) {
	if (flag)
		return;

	if (layer == N) {
		cout << ans[1];
		for (int i = 2; i <= N; i++)
			cout << " " <<  ans[i];
		cout << endl;
		flag = 1;
		return;
	}
	else {
		int vmin = *sum.begin();
		int layer_num = vmin - ans[1];
		ans[layer + 1] = layer_num;
		for (int i = 2; i <= layer; i ++) {
			if (sum.find(layer_num + ans[i]) == sum.end())
				return;
		}

		for (int i = 1; i <= layer; i++) {
			auto iter = sum.find(layer_num + ans[i]);
			sum.erase(iter);
		}

		SolveLayer(layer + 1);

		for (int i = 1; i <= layer; i++) {
			sum.insert(layer_num + ans[i]);
		}
	}
}

void Solve() {
	int v1 = *sum.begin();
	sum.erase(v1);
	int maxv = (v1 - 1) / 2;
	for (int i = 1; i <= maxv; i++) {
		ans[1] = i;
		ans[2] = v1 - i;
		SolveLayer(2);
	}
}

int main() {
	cin >> N;
	int cnt = (N * (N - 1)) / 2;
	for (int i = 1; i <= cnt; i ++) {
		int k;
		cin >> k;
		sum.insert(k);
	}

	Solve();
	return 0;
}