# PROBLEM LINK:

Practice

Contest: Division 1

Contest: Division 2

Contest: Division 3

Contest: Division 4

* Author:* raysh07

*apoorv_me*

**Tester:***iceknight1093*

**Editorialist:**# DIFFICULTY:

TBD

# PREREQUISITES:

None

# PROBLEM:

For an array B of length N, define f(B) = \sum_{i=1}^{N-1} (B_i + B_{i+1}).

You’re given an array A and Q queries on it.

Each query gives you a value X. Find any rearrangement of A such that f(A) = X, or report that no such rearrangement exists.

# EXPLANATION:

It is recommended that you first solve MAXADJSUM (or at least read its editorial, since this one will continue from there).

Recall that after a bit of simplification, we have

where S denotes twice the sum of A.

Now, suppose we want f(A) = X.

This means S - A_1 - A_N = X should hold; or A_1 + A_N = S - X.

In other words, we want two *different* elements of A that add up to S-X.

If you’re able to find such a pair, place one of them at the front of the array, the other one at the back, and everything else inbetween (in any order).

The only thing left is to actually find such a pair.

The constraints are low enough that this can be done with a simple bruteforce algorithm!

That is, simple iterate across all pairs (i, j) (with 1 \leq i \lt j \leq N), and check if A_i + A_j = S-X. Stop once you have a valid pair.

This is \mathcal{O}(N^2) per query, which is fast enough since T, N, Q \leq 50.

Faster solutions exist, but are not necessary to get AC on this problem.

# TIME COMPLEXITY:

\mathcal{O}(N^2 Q) per testcase.

# CODE:

## Author's code (C++)

```
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF (int)1e18
#define f first
#define s second
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
void Solve()
{
int n, q; cin >> n >> q;
vector <int> a(n);
for (auto &x : a) cin >> x;
int sum = accumulate(a.begin(), a.end(), 0);
sum *= 2;
while (q--){
int x; cin >> x;
x = sum - x;
bool print = false;
for (int i = 0; i < n; i++){
for (int j = i + 1; j < n; j++){
if ((a[i] + a[j]) == x && !print){
print = true;
cout << a[i] << " ";
for (int k = 0; k < n; k++) if (k != i && k != j) cout << a[k] << " ";
cout << a[j] << "\n";
}
}
}
if (!print) cout << -1 << "\n";
}
}
int32_t main()
{
auto begin = std::chrono::high_resolution_clock::now();
ios_base::sync_with_stdio(0);
cin.tie(0);
int t = 1;
// freopen("in", "r", stdin);
// freopen("out", "w", stdout);
cin >> t;
for(int i = 1; i <= t; i++)
{
//cout << "Case #" << i << ": ";
Solve();
}
auto end = std::chrono::high_resolution_clock::now();
auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n";
return 0;
}
```

## Editorialist's code (Python)

```
for _ in range(int(input())):
n, q = map(int, input().split())
a = list(map(int, input().split()))
S = 2*sum(a)
for k in range(q):
x = int(input())
p, q = -1, -1
for i in range(n):
for j in range(i+1, n):
if S - a[i] - a[j] == x:
p, q = i, j
if p == -1:
print(-1)
continue
ans = [a[p]] + a[:p] + a[p+1:q] + a[q+1:] + [a[q]]
print(*ans)
```