1. Abridged Problem Statement

Given a spherical planet of radius R centered at the origin and two scanner points outside the sphere, compute the total surface area of the sphere from which at least one scanner is directly visible.

A scanner sees exactly the spherical cap facing it, bounded by the tangent circle from the scanner to the sphere.

Input:
  R
  x1 y1 z1
  x2 y2 z2

Output the visible surface area, rounded to 3 decimal digits.

2. Detailed Editorial

Visibility from one scanner

Let the planet be a sphere of radius R centered at the origin. Suppose a scanner is at point p, with distance from the center d = |p|. Let u = p / d be the unit vector from the center toward the scanner.

A point x on the sphere is visible from the scanner if the segment from x to the scanner does not pass through the planet. The boundary of visibility is the tangent circle from the scanner to the sphere; this tangent circle lies in a plane perpendicular to u. The visible cap is:
  { x on sphere : (x / R) · u >= R / d }

So on the unit sphere, it is a spherical cap with angular radius alpha = acos(R / d).

The area of a spherical cap of angular radius alpha on a sphere of radius R is 2πR²(1 - cos(alpha)). Since cos(alpha) = R / d, the cap area is 2πR²(1 - R / d).

Union of two visible regions

Each scanner sees one spherical cap. Let the two caps have angular radii alpha, beta, and let the angular distance between their centers be theta. Then:
  area(cap1 ∪ cap2) = area(cap1) + area(cap2) - area(cap1 ∩ cap2)

So the key task is computing the intersection area of two spherical caps.

Cases for two caps

Case 1: disjoint caps. If theta >= alpha + beta, the caps do not overlap, so intersection = 0.

Case 2: one cap inside the other. If theta + min(alpha, beta) <= max(alpha, beta), then the smaller cap is fully contained inside the larger cap, so the intersection is just the area of the smaller cap.

Case 3: partial overlap. Otherwise, the intersection is a spherical lens bounded by two small-circle arcs. Let ca = cos(alpha), cb = cos(beta), ct = cos(theta), and sa = sin(alpha), sb = sin(beta), st = sin(theta). Define psi_a as half the angular span of the intersection arc along cap a's boundary (measured around the axis of cap a), and similarly psi_b for cap b. Using spherical trigonometry:
  cos(psi_a) = (cos(beta) - cos(alpha) cos(theta)) / (sin(alpha) sin(theta))
  cos(psi_b) = (cos(alpha) - cos(beta) cos(theta)) / (sin(beta) sin(theta))
Define gamma as the angle between the two boundary tangent directions at an intersection point:
  cos(gamma) = (cos(theta) - cos(alpha) cos(beta)) / (sin(alpha) sin(beta))
By applying Gauss-Bonnet to the spherical lens on the unit sphere, its area is 2(π - gamma) - 2 psi_a cos(alpha) - 2 psi_b cos(beta). For a sphere of radius R, multiply by R²:
  intersection = R² [2(π - gamma) - 2 psi_a cos(alpha) - 2 psi_b cos(beta)]

Finally, answer = area1 + area2 - intersection. All acos arguments should be clamped into [-1, 1] to avoid floating-point errors. Complexity is O(1).

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;
};

using coord3_t = double;

const coord3_t PI = acos((coord3_t)-1.0);

struct Point3 {
    coord3_t x, y, z;

    coord3_t dot(const Point3& p) const { return x * p.x + y * p.y + z * p.z; }
    coord3_t norm() const { return sqrt(dot(*this)); }

    friend istream& operator>>(istream& in, Point3& p) {
        return in >> p.x >> p.y >> p.z;
    }
};

coord3_t R;
Point3 p1, p2;

void read() { cin >> R >> p1 >> p2; }

void solve() {
    // A scanner at distance d from the center sees a spherical cap. The
    // tangent points form a circle lying in the plane perpendicular to the
    // line center-scanner at distance R^2 / d from the center, so in terms
    // of the unit vector u toward the scanner the cap is
    //
    //     { x on sphere : (x / R) . u >= R / d },
    //
    // i.e. a cap of half-angle a = acos(R / d) and area 2 pi R^2 (1 - R/d).
    //
    // By inclusion-exclusion,
    //
    //     answer = area(cap_1) + area(cap_2) - area(cap_1 cap cap_2),
    //
    // so we only need the intersection. On the sphere, the intersection of
    // two caps is a "lens" bounded by two small-circle arcs meeting at the
    // two tangent-circle crossings. Let a, b be the cap half-angles and t
    // the angle between their axes u_a, u_b.
    //
    //   1) Each arc spans a half-angle psi at its pole. Restricting cap a's
    //      boundary (x . u_a = cos a) to also lie in cap b gives
    //
    //          cos(psi_a) = (cos b - cos a cos t) / (sin a sin t),
    //          cos(psi_b) = (cos a - cos b cos t) / (sin b sin t).
    //
    //   2) Each lens corner P is the apex of the spherical triangle
    //      (P, u_a, u_b), so the spherical law of cosines gives the angle g
    //      between the two boundary tangents at P:
    //
    //          cos(g) = (cos t - cos a cos b) / (sin a sin b).
    //
    //      The interior angle of the lens at P is then pi - g.
    //
    //   3) Apply Gauss-Bonnet on the unit sphere (K = 1). A small circle of
    //      half-angle r has geodesic curvature cot r, and an arc spanning
    //      2 psi at its pole has length 2 psi sin r, so the integral of
    //      k_g ds over each arc is 2 psi cos r. With two corners of
    //      interior angle pi - g (i.e. exterior angle g) the theorem reads
    //
    //          area + 2 psi_a cos a + 2 psi_b cos b + 2 g = 2 pi,
    //
    //      hence on a sphere of radius R
    //
    //          area(lens) = R^2 (2(pi - g) - 2 psi_a cos a - 2 psi_b cos b).
    //
    // The disjoint / nested cases are handled explicitly up front; otherwise
    // we clamp the cosines into [-1, 1] before acos for numerical safety.

    coord3_t d1 = p1.norm(), d2 = p2.norm();
    coord3_t cos_a = R / d1, cos_b = R / d2;
    coord3_t sin_a = sqrt(max(0.0, 1.0 - cos_a * cos_a));
    coord3_t sin_b = sqrt(max(0.0, 1.0 - cos_b * cos_b));
    coord3_t alpha = acos(cos_a), beta = acos(cos_b);

    coord3_t cos_t = p1.dot(p2) / (d1 * d2);
    cos_t = max((coord3_t)-1.0, min((coord3_t)1.0, cos_t));
    coord3_t sin_t = sqrt(max(0.0, 1.0 - cos_t * cos_t));
    coord3_t theta = acos(cos_t);

    coord3_t area1 = 2 * PI * R * R * (1 - cos_a);
    coord3_t area2 = 2 * PI * R * R * (1 - cos_b);

    coord3_t intersection = 0.0;
    if(theta >= alpha + beta) {
        intersection = 0.0;
    } else if(theta + min(alpha, beta) <= max(alpha, beta)) {
        intersection = 2 * PI * R * R * (1 - (alpha < beta ? cos_a : cos_b));
    } else {
        auto clamp_cos = [](coord3_t c) {
            return max((coord3_t)-1.0, min((coord3_t)1.0, c));
        };
        coord3_t psi_a =
            acos(clamp_cos((cos_b - cos_a * cos_t) / (sin_a * sin_t)));
        coord3_t psi_b =
            acos(clamp_cos((cos_a - cos_b * cos_t) / (sin_b * sin_t)));
        coord3_t gamma =
            acos(clamp_cos((cos_t - cos_a * cos_b) / (sin_a * sin_b)));
        intersection =
            R * R * (2 * (PI - gamma) - 2 * psi_a * cos_a - 2 * psi_b * cos_b);
    }

    cout << fixed << setprecision(3) << area1 + area2 - intersection << '\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

```python
import math
import sys


def clamp(x, lo=-1.0, hi=1.0):
    """
    Clamp x into the interval [lo, hi].

    This is important before calling acos(), because floating-point
    computations can produce values such as 1.0000000000000002.
    """
    return max(lo, min(hi, x))


def dot(a, b):
    """
    Dot product of two 3D vectors.
    """
    return a[0] * b[0] + a[1] * b[1] + a[2] * b[2]


def norm(a):
    """
    Euclidean length of a 3D vector.
    """
    return math.sqrt(dot(a, a))


def main():
    # Read all input numbers.
    data = sys.stdin.read().strip().split()

    # If input is empty, do nothing.
    if not data:
        return

    # First value is the radius of the planet.
    R = float(data[0])

    # Next three values are coordinates of scanner 1.
    p1 = (
        float(data[1]),
        float(data[2]),
        float(data[3]),
    )

    # Last three values are coordinates of scanner 2.
    p2 = (
        float(data[4]),
        float(data[5]),
        float(data[6]),
    )

    # Mathematical constant pi.
    pi = math.pi

    # Distances from the center of the planet to each scanner.
    d1 = norm(p1)
    d2 = norm(p2)

    # For a scanner at distance d, the visible spherical cap has
    # angular radius alpha satisfying cos(alpha) = R / d.
    cos_a = R / d1
    cos_b = R / d2

    # Compute the corresponding sines safely.
    sin_a = math.sqrt(max(0.0, 1.0 - cos_a * cos_a))
    sin_b = math.sqrt(max(0.0, 1.0 - cos_b * cos_b))

    # Angular radii of the two visible caps.
    alpha = math.acos(clamp(cos_a))
    beta = math.acos(clamp(cos_b))

    # Angle theta between scanner direction vectors.
    cos_t = dot(p1, p2) / (d1 * d2)
    cos_t = clamp(cos_t)

    # sin(theta), computed safely.
    sin_t = math.sqrt(max(0.0, 1.0 - cos_t * cos_t))

    # Actual angle theta.
    theta = math.acos(cos_t)

    # Area of each spherical cap:
    # area = 2πR²(1 - cos(radius_angle)).
    area1 = 2.0 * pi * R * R * (1.0 - cos_a)
    area2 = 2.0 * pi * R * R * (1.0 - cos_b)

    # Compute intersection area of the two caps.
    if theta >= alpha + beta:
        # Caps are disjoint or tangent.
        intersection = 0.0

    elif theta + min(alpha, beta) <= max(alpha, beta):
        # One cap lies completely inside the other.
        # Intersection equals the smaller cap.
        if alpha < beta:
            intersection = area1
        else:
            intersection = area2

    else:
        # Partial overlap: the intersection is a spherical lens.

        # Half-angle span of the arc from cap 1's boundary.
        psi_a = math.acos(
            clamp((cos_b - cos_a * cos_t) / (sin_a * sin_t))
        )

        # Half-angle span of the arc from cap 2's boundary.
        psi_b = math.acos(
            clamp((cos_a - cos_b * cos_t) / (sin_b * sin_t))
        )

        # Angle between boundary tangent directions at an intersection point.
        gamma = math.acos(
            clamp((cos_t - cos_a * cos_b) / (sin_a * sin_b))
        )

        # Lens area from Gauss-Bonnet on the unit sphere,
        # then scaled by R².
        intersection = R * R * (
            2.0 * (pi - gamma)
            - 2.0 * psi_a * cos_a
            - 2.0 * psi_b * cos_b
        )

    # Union area by inclusion-exclusion.
    answer = area1 + area2 - intersection

    # Print rounded to 3 digits after the decimal point.
    print(f"{answer:.3f}")


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

5. Compressed Editorial

Each scanner sees a spherical cap. If the scanner is at distance d from the sphere center, then the cap angular radius a satisfies cos(a) = R / d, and its area is 2πR²(1 - R / d).

Let the two caps have angular radii a, b, and let the angle between scanner direction vectors be t. The answer is area1 + area2 - intersection.

If t >= a + b, then intersection = 0. If t + min(a, b) <= max(a, b), then one cap is inside the other, and intersection is the smaller cap area. Otherwise, the intersection is a spherical lens. Let ca = cos(a), cb = cos(b), ct = cos(t), sa = sin(a), sb = sin(b), st = sin(t). Then:
  psi_a = acos((cb - ca ct) / (sa st))
  psi_b = acos((ca - cb ct) / (sb st))
  gamma = acos((ct - ca cb) / (sa sb))
The lens area is R² * (2(π - gamma) - 2 psi_a ca - 2 psi_b cb). Clamp all acos arguments to [-1, 1]. The solution is constant time, O(1).
