## 1) Abridged problem statement

Given \(N\) points (\(3 \le N \le 8\)) on a plane, each point \(i\) has a positive “special number” \(R_i\) (\(1 \le R_i \le 100\)), and it is guaranteed that \(R_{N-1} = R_N\) (last two are equal).

An arrangement of points is called **awful** if for **every** triple of distinct points \((i,j,k)\), the **area** of triangle \(\triangle ijk\) equals:
\[
\text{Area}(i,j,k) = R_i + R_j + R_k.
\]

You must output **YES** and any coordinates \((x_i, y_i)\) (4 decimals) that satisfy all constraints, or **NO** if impossible.

---

## 2) Detailed editorial (explaining the provided solution approach)

### Key geometry identity

For three points \((x_i,y_i)\), \((x_j,y_j)\), \((x_k,y_k)\), the **signed doubled area** is the 2D cross product:
\[
2 \cdot \text{Area} = \left| (x_i-x_j)(y_k-y_j) - (x_k-x_j)(y_i-y_j) \right|.
\]

Equivalently, expanding (a common determinant form):
\[
\text{cross}(i,j,k) = x_i(y_k-y_j) + x_j(y_i-y_k) + x_k(y_j-y_i),
\]
and
\[
2\cdot \text{Area} = |\text{cross}(i,j,k)|.
\]

The requirement \(\text{Area} = R_i+R_j+R_k\) becomes:
\[
|\text{cross}(i,j,k)| = 2(R_i+R_j+R_k).
\]

### Removing the absolute value

The absolute value causes branching (two possible signs). The code avoids it by squaring:
\[
\text{cross}^2 = \left(2(R_i+R_j+R_k)\right)^2.
\]

Define:
- \(T = 2(R_i+R_j+R_k)\)
- \(f = \text{cross}^2 - T^2\)

We want \(f=0\) for all triples. There are \(\binom{N}{3}\) triples (at most \(\binom{8}{3}=56\)), so this is small.

### Turn it into optimization (least squares)

Instead of solving polynomial equations directly, the solution minimizes:
\[
\text{loss} = \sum_{i<j<k} f(i,j,k)^2.
\]
If loss becomes (near) 0, we have a valid configuration.

This transforms the problem into a continuous optimization over \(2N\) variables:
\[
(x_0,\dots,x_{N-1}, y_0,\dots,y_{N-1}).
\]

### Gradient computation

For one triple, with:
\[
f = \text{cross}^2 - T^2,
\quad \text{loss term} = f^2,
\]
the derivative is:
\[
\frac{\partial}{\partial p}(f^2) = 2f \cdot \frac{\partial f}{\partial p}
= 2f \cdot 2\cdot \text{cross}\cdot \frac{\partial \text{cross}}{\partial p}
= 4 f\cdot \text{cross}\cdot \frac{\partial \text{cross}}{\partial p}.
\]

Let \(\text{coef} = 4 f\cdot \text{cross}\).

Now, from
\[
\text{cross} = x_i(y_k-y_j) + x_j(y_i-y_k) + x_k(y_j-y_i),
\]
we get partials:
- \(\partial \text{cross}/\partial x_i = (y_k-y_j)\)
- \(\partial \text{cross}/\partial x_j = (y_i-y_k)\)
- \(\partial \text{cross}/\partial x_k = (y_j-y_i)\)

And for y’s (collect coefficients of \(y_i,y_j,y_k\)):
- \(\partial \text{cross}/\partial y_i = (x_j - x_k)\)
- \(\partial \text{cross}/\partial y_j = (x_k - x_i)\)
- \(\partial \text{cross}/\partial y_k = (x_i - x_j)\)

So each triple contributes to gradients of exactly 6 variables.

### Optimizer: Adam + random restarts

The loss is a high-degree polynomial and non-convex, so local minima exist. The code uses:
- **Random initialization** of all points in a square \([-15,15]\times[-15,15]\)
- **Adam** optimizer (adaptive moments), which is robust for poorly scaled problems
- **Multiple restarts** (up to 80) to escape bad basins

Stopping/acceptance:
- If a restart achieves extremely tiny loss, stop early.
- After all restarts, if best loss is below a threshold (\(1e{-4}\)), print **YES** and coordinates; else **NO**.

### Complexity

For each gradient evaluation:
- \(O(\binom{N}{3})\) triples, each O(1) work
- With \(N\le 8\), max 56 triples
So even a few thousand steps and many restarts can fit, though the approach is heuristic.

### Important note

This is a **numerical / heuristic** approach, not a guaranteed constructive proof. It often works for such small \(N\) but is not theoretically certain under all judges. (The original problem likely expects an exact constructive method; the provided solution instead “searches” for coordinates.)

---

## 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;
vector<double> r_vals;

void read() {
    cin >> n;
    r_vals.assign(n, 0.0);
    cin >> r_vals;
}

void solve() {
    // The constraint for N is fairly small. We can try a numerical approach -
    // let's create a system for this, with 2*n variables (x[:], y[:]). We then
    // have C(n, 3) constraints, that are given by:
    //
    //     v[i] + v[j] + v[k] = abs((x[i] - x[j]) * (y[k] - y[j]) -
    //                              (x[k] - x[j]) * (y[i] - y[j])).
    //
    // By squaring both sides, we get rid of the absolute value and the
    // branching.
    //
    //     Const = a((x[i] - x[j]) * (y[k] - y[j]) -
    //               (x[k] - x[j]) * (y[i] - y[j]))^2
    //
    // We can now solve this with gradient descent by instead minimizing the sum
    // of squares. The term itself will be of high order as we have products of
    // variables and then we ^4, but it's still a polynomial. After some
    // tuning, we can get a solution that runs in a reasonable amount of time.
    // Here I implement Adam with random restarts, and do the gradient with
    // chain rule as it's easier.

    vector<double> x(n), y(n), gx(n), gy(n);
    vector<double> m_x(n), m_y(n), v_x(n), v_y(n);

    mt19937 rng(987654321);
    uniform_real_distribution<double> init_dist(-15.0, 15.0);

    auto compute = [&](bool need_grad) -> double {
        if(need_grad) {
            fill(gx.begin(), gx.end(), 0.0);
            fill(gy.begin(), gy.end(), 0.0);
        }
        double loss = 0.0;
        for(int i = 0; i < n; i++) {
            for(int j = i + 1; j < n; j++) {
                for(int k = j + 1; k < n; k++) {
                    double cross = x[i] * (y[k] - y[j]) + x[j] * (y[i] - y[k]) +
                                   x[k] * (y[j] - y[i]);
                    double target = 2.0 * (r_vals[i] + r_vals[j] + r_vals[k]);
                    double f = cross * cross - target * target;
                    loss += f * f;
                    if(need_grad) {
                        double coef = 4.0 * f * cross;
                        gx[i] += coef * (y[k] - y[j]);
                        gx[j] += coef * (y[i] - y[k]);
                        gx[k] += coef * (y[j] - y[i]);
                        gy[i] += coef * (x[j] - x[k]);
                        gy[j] += coef * (x[k] - x[i]);
                        gy[k] += coef * (x[i] - x[j]);
                    }
                }
            }
        }
        return loss;
    };

    double best_loss = numeric_limits<double>::infinity();
    vector<double> best_x(n), best_y(n);

    const double beta1 = 0.9, beta2 = 0.999, adam_eps = 1e-8;
    const double lr = 0.1;
    const int max_restarts = 80;
    const int max_steps = 2500;

    for(int restart = 0; restart < max_restarts; restart++) {
        for(int i = 0; i < n; i++) {
            x[i] = init_dist(rng);
            y[i] = init_dist(rng);
        }
        fill(m_x.begin(), m_x.end(), 0.0);
        fill(m_y.begin(), m_y.end(), 0.0);
        fill(v_x.begin(), v_x.end(), 0.0);
        fill(v_y.begin(), v_y.end(), 0.0);

        double b1t = 1.0, b2t = 1.0;
        for(int step = 1; step <= max_steps; step++) {
            double loss = compute(true);
            if(loss < 1e-20) {
                break;
            }
            b1t *= beta1;
            b2t *= beta2;
            double bc1 = 1.0 - b1t;
            double bc2 = 1.0 - b2t;
            for(int i = 0; i < n; i++) {
                m_x[i] = beta1 * m_x[i] + (1.0 - beta1) * gx[i];
                m_y[i] = beta1 * m_y[i] + (1.0 - beta1) * gy[i];
                v_x[i] = beta2 * v_x[i] + (1.0 - beta2) * gx[i] * gx[i];
                v_y[i] = beta2 * v_y[i] + (1.0 - beta2) * gy[i] * gy[i];
                x[i] -= lr * (m_x[i] / bc1) / (sqrt(v_x[i] / bc2) + adam_eps);
                y[i] -= lr * (m_y[i] / bc1) / (sqrt(v_y[i] / bc2) + adam_eps);
            }
        }

        double final_loss = compute(false);
        if(final_loss < best_loss) {
            best_loss = final_loss;
            best_x = x;
            best_y = y;
        }
        if(best_loss < 1e-12) {
            break;
        }
    }

    if(best_loss < 1e-4) {
        cout << "YES\n";
        cout << fixed << setprecision(4);
        for(int i = 0; i < n; i++) {
            cout << best_x[i] << ' ' << best_y[i] << '\n';
        }
    } else {
        cout << "NO\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();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same numerical method) with detailed comments

```python
import sys
import math
import random

def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    r = [float(next(it)) for _ in range(n)]

    # Variables: x[i], y[i]
    x = [0.0] * n
    y = [0.0] * n

    # Gradients
    gx = [0.0] * n
    gy = [0.0] * n

    # Adam state
    mx = [0.0] * n
    my = [0.0] * n
    vx = [0.0] * n
    vy = [0.0] * n

    rng = random.Random(987654321)

    def compute(need_grad: bool) -> float:
        """Compute total loss; if need_grad, also compute gradients into gx,gy."""
        if need_grad:
            for i in range(n):
                gx[i] = 0.0
                gy[i] = 0.0

        loss = 0.0
        # All triples i < j < k
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    # cross is doubled signed area:
                    # cross = x_i(y_k-y_j) + x_j(y_i-y_k) + x_k(y_j-y_i)
                    cross = (
                        x[i] * (y[k] - y[j]) +
                        x[j] * (y[i] - y[k]) +
                        x[k] * (y[j] - y[i])
                    )

                    # Target doubled area
                    target = 2.0 * (r[i] + r[j] + r[k])

                    # Residual after squaring to remove abs:
                    # want cross^2 = target^2
                    f = cross * cross - target * target
                    loss += f * f

                    if need_grad:
                        # d(f^2)/d(p) = 4*f*cross * d(cross)/d(p)
                        coef = 4.0 * f * cross

                        # d(cross)/d(x_i) etc.
                        gx[i] += coef * (y[k] - y[j])
                        gx[j] += coef * (y[i] - y[k])
                        gx[k] += coef * (y[j] - y[i])

                        # d(cross)/d(y_i) etc.
                        gy[i] += coef * (x[j] - x[k])
                        gy[j] += coef * (x[k] - x[i])
                        gy[k] += coef * (x[i] - x[j])

        return loss

    # Hyperparameters (mirrors the C++ code)
    beta1 = 0.9
    beta2 = 0.999
    eps = 1e-8
    lr = 0.1
    max_restarts = 80
    max_steps = 2500

    best_loss = float("inf")
    best_x = None
    best_y = None

    for _ in range(max_restarts):
        # Random init in [-15,15]
        for i in range(n):
            x[i] = rng.uniform(-15.0, 15.0)
            y[i] = rng.uniform(-15.0, 15.0)

        # Reset Adam
        for i in range(n):
            mx[i] = my[i] = 0.0
            vx[i] = vy[i] = 0.0

        b1t = 1.0  # beta1^t
        b2t = 1.0  # beta2^t

        for _step in range(1, max_steps + 1):
            loss = compute(True)

            # Converged extremely well
            if loss < 1e-20:
                break

            # Update bias-correction powers
            b1t *= beta1
            b2t *= beta2
            bc1 = 1.0 - b1t
            bc2 = 1.0 - b2t

            # Adam parameter update
            for i in range(n):
                mx[i] = beta1 * mx[i] + (1.0 - beta1) * gx[i]
                my[i] = beta1 * my[i] + (1.0 - beta1) * gy[i]

                vx[i] = beta2 * vx[i] + (1.0 - beta2) * gx[i] * gx[i]
                vy[i] = beta2 * vy[i] + (1.0 - beta2) * gy[i] * gy[i]

                # Bias-corrected moments
                mhatx = mx[i] / bc1
                mhaty = my[i] / bc1
                vhatx = vx[i] / bc2
                vhaty = vy[i] / bc2

                x[i] -= lr * mhatx / (math.sqrt(vhatx) + eps)
                y[i] -= lr * mhaty / (math.sqrt(vhaty) + eps)

        final_loss = compute(False)
        if final_loss < best_loss:
            best_loss = final_loss
            best_x = x[:]
            best_y = y[:]

        if best_loss < 1e-12:
            break

    if best_loss < 1e-4:
        out = ["YES"]
        for i in range(n):
            out.append(f"{best_x[i]:.4f} {best_y[i]:.4f}")
        sys.stdout.write("\n".join(out) + "\n")
    else:
        sys.stdout.write("NO\n")


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

---

## 5) Compressed editorial

We need coordinates so that for every triple \((i,j,k)\),
\[
\text{Area}(i,j,k)=R_i+R_j+R_k.
\]
Using doubled area via cross product:
\[
|\text{cross}(i,j,k)| = 2(R_i+R_j+R_k),
\]
square to remove absolute:
\[
\text{cross}^2 - \text{target}^2 = 0,\quad \text{target}=2(R_i+R_j+R_k).
\]
Minimize least-squares loss:
\[
\sum_{i<j<k}(\text{cross}^2-\text{target}^2)^2
\]
over variables \((x_i,y_i)\). Compute gradients analytically:
\(\nabla = 4 f \cdot \text{cross} \cdot \nabla(\text{cross})\).
Run Adam optimizer for a few thousand steps with multiple random restarts; if best loss is small enough, output coordinates, otherwise output NO.