<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

251. Polymania
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



Consider N points on the plane. Each point has its special number. Points arrangement is called "awful" if for any three different points A, B, and C the sum of their special numbers is equal to the area of the triangle ABC.
You are given special numbers for all points, but not their coordinates. You have to find some awful points arrangement.
It is guarantee, that last special number and previous are equal.

Input
The first line of input contains single integer N (3<=N<=8). Next line contains N positive integers R[i] - special numbers of points (0<R[i]<=100, R[N-1]=R[N]).

Output
On the first line output "YES" (without quotes) if the solution exists, or "NO" (without quotes) if it does not. In case of positive answer output N lines with points coordinates: X[i] and Y[i] with four digits after the decimal point.

Sample test(s)

Input
Test #1
3
2 3 3

Test #2
4
1 1 1 1

Output
Test #1
YES
1.5000 0.0000
5.5000 0.0000
1.5000 4.0000

Test #2
YES
0.0000 0.0000
0.0000 2.4495
2.4495 0.0000
2.4495 2.4495
Author:	Alexey Preobrajensky
Resource:	Petrozavodsk Summer Training Sessions 2004
Date:	August 25, 2004

<|response|>
## 1) Abridged problem statement

You are given \(N\) (\(3 \le N \le 8\)) positive integers \(R_1,\dots,R_N\) (“special numbers”), with a guarantee \(R_{N-1}=R_N\).  
Find coordinates \((x_i,y_i)\) for \(N\) points such that for **every** triple of distinct points \((i,j,k)\),

\[
\text{Area}(\triangle ijk) = R_i + R_j + R_k.
\]

Output **YES** and any valid coordinates (4 decimals) if possible, otherwise output **NO**.

---

## 2) Key observations needed to solve the problem

1. **Triangle area via cross product (doubled area):**  
   For points \(i,j,k\),
   \[
   2\cdot \text{Area} = \left| \text{cross}(i,j,k) \right|
   \]
   where
   \[
   \text{cross}(i,j,k)=x_i(y_k-y_j)+x_j(y_i-y_k)+x_k(y_j-y_i).
   \]

2. **Turn constraints into equations without absolute value:**  
   Required:
   \[
   |\text{cross}(i,j,k)| = 2(R_i+R_j+R_k).
   \]
   Remove \(|\cdot|\) by squaring:
   \[
   \text{cross}(i,j,k)^2 = \left(2(R_i+R_j+R_k)\right)^2.
   \]

3. **Solve as nonlinear least squares (numerical):**  
   For each triple define residual
   \[
   f_{ijk} = \text{cross}^2 - T^2,\quad T=2(R_i+R_j+R_k).
   \]
   Minimize
   \[
   \text{loss}=\sum_{i<j<k} f_{ijk}^2
   \]
   over \(2N\) real variables \(\{x_i,y_i\}\).

4. **Analytical gradients are easy and fast (small \(N\)):**  
   With \(N\le 8\), there are at most \(\binom{8}{3}=56\) triples. We can compute loss+gradient quickly and run an optimizer with random restarts.

5. **Heuristic approach:**  
   This is a numerical method (Adam + restarts). It works well in practice for this tiny size, but it’s not a symbolic/guaranteed constructive proof.

---

## 3) Full solution approach based on the observations

### Step A — Build the equation for one triple
For each triple \((i,j,k)\):

- Compute doubled signed area:
  \[
  c = \text{cross}(i,j,k)
  \]
- Target doubled area:
  \[
  T = 2(R_i+R_j+R_k)
  \]
- Residual:
  \[
  f = c^2 - T^2
  \]
We want \(f=0\) for all triples.

### Step B — Define the objective (loss)
\[
\text{loss} = \sum_{i<j<k} f(i,j,k)^2.
\]
If loss is ~0, all constraints are satisfied (up to numeric error).

### Step C — Compute gradients
For one triple,
\[
\frac{\partial}{\partial p}(f^2)=2f\cdot \frac{\partial f}{\partial p}
=2f\cdot 2c\cdot \frac{\partial c}{\partial p}
=4fc\cdot \frac{\partial c}{\partial p}.
\]
Let \(\text{coef}=4fc\).

From
\[
c=x_i(y_k-y_j)+x_j(y_i-y_k)+x_k(y_j-y_i),
\]
the partial derivatives are:
- w.r.t. x:
  - \(\partial c/\partial x_i = (y_k-y_j)\)
  - \(\partial c/\partial x_j = (y_i-y_k)\)
  - \(\partial c/\partial x_k = (y_j-y_i)\)
- w.r.t. y:
  - \(\partial c/\partial y_i = (x_j-x_k)\)
  - \(\partial c/\partial y_j = (x_k-x_i)\)
  - \(\partial c/\partial y_k = (x_i-x_j)\)

Each triple updates gradients of exactly 6 variables.

### Step D — Minimize loss with Adam + random restarts
Because the loss is non-convex, we:
- Randomly initialize all points in some box (e.g. \([-15,15]\)).
- Run Adam for a fixed number of steps.
- Repeat several restarts and keep the best.

### Step E — Decide YES/NO
- If best loss < small threshold (e.g. \(10^{-4}\)), print **YES** and coordinates.
- Otherwise print **NO**.

---

## 4) C++ implementation with detailed comments

```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;
}
```

---

## 5) Python implementation 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: coordinates
    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:
        """Return total loss. If need_grad=True, also fills gx, gy with gradients."""
        if need_grad:
            for i in range(n):
                gx[i] = 0.0
                gy[i] = 0.0

        loss = 0.0
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    # doubled signed area (cross product)
                    cross = (
                        x[i] * (y[k] - y[j]) +
                        x[j] * (y[i] - y[k]) +
                        x[k] * (y[j] - y[i])
                    )
                    target = 2.0 * (R[i] + R[j] + R[k])

                    # residual after squaring (removes abs)
                    f = cross * cross - target * target
                    loss += f * f

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

                        # d(cross)/d(x_*)
                        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_*)
                        gy[i] += coef * (x[j] - x[k])
                        gy[j] += coef * (x[k] - x[i])
                        gy[k] += coef * (x[i] - x[j])

        return loss

    # Adam hyperparameters
    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
        b2t = 1.0

        for _step in range(1, max_steps + 1):
            loss = compute(True)
            if loss < 1e-20:
                break

            b1t *= beta1
            b2t *= beta2
            bc1 = 1.0 - b1t
            bc2 = 1.0 - b2t

            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

                # Adam update
                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()
```

--- 

If you want a more “judge-safe” approach, I can try to derive a deterministic constructive method (if one exists for this problem) or add a final verification pass that checks all triangle areas and, if needed, retries with different random seeds.