## 1) Abridged problem statement

Given two infinite right circular cylinders in 3D with radii \(R_1\) and \(R_2\). Their axes intersect at a point and are perpendicular to each other. Compute the volume of the solid that lies inside **both** cylinders.  
Input: two real numbers \(R_1, R_2\) (1…100).  
Output: intersection volume accurate to \(10^{-4}\).

---

## 2) Detailed editorial (solution explanation)

### Geometry setup
Place the origin at the intersection point of the two cylinder axes.

- Cylinder 1 (radius \(r\)) axis along the **x-axis**:
  \[
  y^2 + z^2 \le r^2
  \]
- Cylinder 2 (radius \(R\)) axis along the **y-axis**:
  \[
  x^2 + z^2 \le R^2
  \]
WLOG let \(r = \min(R_1,R_2)\), \(R = \max(R_1,R_2)\) (the code enforces this by swapping if needed).

### Cross-section at fixed \(z\)
For a fixed \(z\):

From cylinder 1:
\[
y^2 \le r^2 - z^2 \Rightarrow |y| \le \sqrt{r^2 - z^2}
\]

From cylinder 2:
\[
x^2 \le R^2 - z^2 \Rightarrow |x| \le \sqrt{R^2 - z^2}
\]

So at that \(z\), the allowed \((x,y)\) region is a rectangle:
- width in \(x\): \(2\sqrt{R^2 - z^2}\)
- width in \(y\): \(2\sqrt{r^2 - z^2}\)

Area:
\[
A(z) = 4\sqrt{(r^2 - z^2)(R^2 - z^2)}
\]

Also, the common constraint in \(z\) comes from the smaller radius cylinder:
\[
|z| \le r
\]

### Volume integral
\[
V = \int_{-r}^{r} A(z)\,dz
  = \int_{-r}^{r} 4\sqrt{(r^2 - z^2)(R^2 - z^2)}\,dz
\]

The integrand is even in \(z\), so:
\[
V = 8\int_{0}^{r}\sqrt{(r^2 - z^2)(R^2 - z^2)}\,dz
\]

Define:
\[
f(z)=\sqrt{(r^2 - z^2)(R^2 - z^2)}
\]
Then:
\[
V = 8\int_0^r f(z)\,dz
\]

### Numerical integration: Simpson’s rule
The integral has a smooth integrand on \([0,r]\), so we can approximate it accurately using Simpson’s rule.

Split \([0,r]\) into \(n\) equal parts (where \(n\) is even), step \(h = r/n\). Then:
\[
\int_0^r f(z)\,dz \approx \frac{h}{3}
\left(
f(0)+f(r)+4\sum_{i\ \text{odd}} f(ih) + 2\sum_{i\ \text{even}, i\neq 0,n} f(ih)
\right)
\]

Finally:
\[
V \approx 8 \cdot \left(\text{Simpson approximation}\right)
\]

The provided solution uses \(n=1{,}000{,}000\), which is easily fast enough for one test case and yields the required \(10^{-4}\) precision.

*(Note: there is also a closed-form using complete elliptic integrals, but it’s unnecessary here.)*

---

## 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;
};

double r1, r2;

void read() { cin >> r1 >> r2; }

void solve() {
    // Let's have r = min(r1, r2), and R = max(r1, r2). The rough idea is that
    // we can put the origin of the coordinate system where the two axes
    // intersect. Let's place the axis of the first cylinder (radius r) along
    // the x-axis, and the second (radius R) along the y-axis. Then we are
    // interested in points x,y,z satisfying:
    //
    //     y^2 + z^2 <= r^2
    //     x^2 + z^2 <= R^2
    //
    // The z coordinate is common to both, and it's clear that |z| <= r. We also
    // have the symmetry between z < 0 and z > 0, so we can just integrate over
    // z from 0 to r. For a fixed z, we have:
    //
    //     |y| <= sqrt(r^2 - z^2)
    //     |x| <= sqrt(R^2 - z^2)
    //
    // We again have the 2x2 symmetries here, but otherwise we have a
    // rectangular range of sqrt(r^2 - z^2) x sqrt(R^2 - z^2). Putting
    // everything together, and having the 2x2x2 = 8 symmetries on the front, we
    // simply want to integrate the below:
    //
    //     V = 8 * integral from 0 to r of sqrt((r^2 - z^2)(R^2 - z^2)) dz
    //
    // Afterwards, the integral can be numerically approximated using Simpson's
    // rule or some other approach, or we can go further by evaluating it to the
    // closed form expression involving complete elliptic integrals of the first
    // kind K(.) and second kind E(.). Then we can evaluate it quicker. The full
    // closed form would be:
    //
    //     V = 8/3 R^3 [ (1 + (r/R)^2) E(r/R) - (1 - (r/R)^2) K(r/R)]
    //
    // However, in this problem we have a single test case so going for
    // Simpson's rule is totally fine. As a reminder, Simpson's approximation is
    // essentially choosing N points at equal distances, and then interpolating
    // a quadratic based on f(l), f((l+r)/2), and f(r). After integrating the
    // quadratic, the approximation is that:
    //
    // integral from l to r f(x) dx = [(r - l) / 6] (f(l) + 4 f((l+r)/2) + f(r))
    //
    // In this problem 1M operations are plenty.

    if(r1 > r2) {
        swap(r1, r2);
    }

    auto f = [](double y) {
        return sqrt((r1 * r1 - y * y) * (r2 * r2 - y * y));
    };

    int n = 1000000;
    double h = r1 / n;
    double sum = f(0) + f(r1);

    for(int i = 1; i < n; i++) {
        double y = i * h;
        if(i % 2 == 1) {
            sum += 4 * f(y);
        } else {
            sum += 2 * f(y);
        }
    }

    double vol = 8 * sum * h / 3;
    cout << fixed << setprecision(4) << vol << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
import math

def main():
    # Read two floating-point radii
    data = sys.stdin.read().strip().split()
    r1 = float(data[0])
    r2 = float(data[1])

    # Ensure r1 <= r2 (r1 is the smaller radius r, r2 is the larger radius R)
    if r1 > r2:
        r1, r2 = r2, r1

    # Integrand f(z) = sqrt((r^2 - z^2)(R^2 - z^2))
    def f(z: float) -> float:
        return math.sqrt((r1*r1 - z*z) * (r2*r2 - z*z))

    # Simpson's rule parameters.
    # n must be even; large n gives high accuracy for required 1e-4 output.
    n = 1_000_000
    h = r1 / n  # step size on [0, r1]

    # Simpson sum initialization with endpoints
    s = f(0.0) + f(r1)

    # Add interior points with alternating weights 4 and 2
    for i in range(1, n):
        z = i * h
        if i % 2 == 1:
            s += 4.0 * f(z)
        else:
            s += 2.0 * f(z)

    # Approximate integral_0^{r1} f(z) dz
    integral = s * h / 3.0

    # Total volume includes factor 8 from symmetry
    volume = 8.0 * integral

    # Print with 4 digits after decimal point
    sys.stdout.write(f"{volume:.4f}\n")

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

---

## 5) Compressed editorial

Let \(r=\min(R_1,R_2)\), \(R=\max(R_1,R_2)\). Place axes perpendicular through the origin, cylinder of radius \(r\) along x: \(y^2+z^2\le r^2\), radius \(R\) along y: \(x^2+z^2\le R^2\).  
For fixed \(z\in[0,r]\), allowed \(x,y\) ranges are \(|x|\le\sqrt{R^2-z^2}\), \(|y|\le\sqrt{r^2-z^2}\), giving cross-sectional area \(4\sqrt{(r^2-z^2)(R^2-z^2)}\). By symmetry in \(x,y,z\),
\[
V=8\int_0^r \sqrt{(r^2-z^2)(R^2-z^2)}\,dz.
\]
Compute the integral numerically via Simpson’s rule on \([0,r]\) with large even \(n\) (e.g. \(10^6\)), then output \(V\) with 4 decimals.