## 1) Abridged problem statement

A beetle moves along the straight segment from \((x_1,y_1)\) to \((x_2,y_2)\) on an infinite grid of unit square cells whose borders lie on integer coordinates. A cell counts as “passed” if the beetle’s path goes through its **interior** (borders/corners do **not** belong to any cell). List all such cells in order along the movement direction, and output the integer coordinates \((X,Y)\) of the **lower-left corner** of the \(n\)-th visited cell. If fewer than \(n\) cells are visited, output `no solution`.

Input: integers \(x_1,y_1,x_2,y_2,n\), with \(1 \le n \le 10^5\), \(|x_i|,|y_i|\le 10^6\).

---

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

### Geometry model of a “cell visited”
Each grid cell can be identified by integer coordinates \((c_x,c_y)\) meaning the open square:
\[
(c_x, c_x+1) \times (c_y, c_y+1)
\]
The beetle travels along the segment:
\[
P(t) = A + (B-A)t,\quad t\in[0,1]
\]
A cell is “passed” iff the segment intersects the **open** square (interior only). Touching only the border should not count.

So for each candidate cell we need to decide if there exists \(t\in(0,1)\) such that:
\[
c_x < x(t) < c_x+1,\qquad c_y < y(t) < c_y+1
\]

### Key idea in code: gather candidate cells cheaply, then sort by entry time
A classical exact approach would be a grid traversal (2D DDA / Amanatides–Woo) in \(O(\#cells)\). The provided code instead does:

1. Sample points along the segment at step length ~1 in Euclidean distance:
   - Let `len = |B-A|`, step in parameter space is `dt = 1/len`.
   - Iterate `t = 0, dt, 2dt, ... , 1`.

2. For each sampled point, compute the cell containing it via `floor(x), floor(y)`.
3. Add that cell and its 8 neighbors (3×3 block) to a list.
   - Why neighbors? If the segment crosses near a cell boundary between samples, the true crossed cell could be adjacent to the sampled cell. The neighborhood compensates for this coarse sampling.
4. For each candidate cell, verify actual intersection using an exact-ish integer slab test (`intersects_cell`) so false candidates are filtered out.
5. Sort the remaining cells by their “entry parameter” (the earliest \(t\) where the line enters the cell’s bounding box), and unique them.
6. Output the \(n\)-th cell in that sorted order, or `no solution`.

Given constraints and a very small time limit, this works because:
- Segment length is at most about \( \sqrt{(2\cdot 10^6)^2+(2\cdot 10^6)^2} \approx 2.8\cdot 10^6\).
- But we only need up to about `8*n` candidates; the loop stops once it collected enough.
- With \(n \le 10^5\), candidate size is capped around \(8\cdot 10^5\), manageable.

### Intersection test for a cell (important part)

Let:
- \(x(t) = x_1 + dx \cdot t\), where `dx = x2-x1` using integer-cast endpoints in the code.
- Similar for y.

For the segment to intersect the *open* cell, the set of \(t\) such that \(x(t)\in(c_x, c_x+1)\) is an interval; same for y. Intersect those intervals with \([0,1]\). If intersection is non-empty (open), the segment passes through interior.

The code implements a “slab intersection”:
- Maintain a rational interval \((t_{\min}, t_{\max})\) as fractions `num/den` to reduce floating issues.
- For x dimension:
  - Compute \(t\) values when line hits vertical boundaries \(x=c_x\) and \(x=c_x+1\):
    \[
    t_1 = \frac{c_x - x_1}{dx},\quad t_2 = \frac{c_x+1 - x_1}{dx}
    \]
  - Take the min/max of these as the interval where x is inside the strip.
  - Combine with current \([t_{\min}, t_{\max}]\).
- If `dx == 0`, the segment is vertical; then it intersects the cell only if `x1` lies strictly inside \((c_x, c_x+1)\). The code checks the negation and returns false.
- Repeat for y.
- Finally it returns `t_min < t_max` using cross-multiplication:
  ```
  return t_min_num * t_max_den < t_max_num * t_min_den;
  ```
Strict `<` (not `<=`) corresponds to open interior (excluding boundaries).

### Ordering cells: “entry time”
To output the n-th visited cell, we need correct order along the segment.
For each cell, define its entry parameter as:
\[
t_\text{entry} = \max( \min(t_{x1}, t_{x2}),\ \min(t_{y1}, t_{y2}),\ 0 )
\]
This is the earliest parameter where the line is inside both x and y slabs (i.e., enters the cell’s bounding box). The function `get_entry_t` computes this in doubles (used only for sorting, after intersection is already validated).

After sorting, duplicates are removed with `unique`. (Note: `unique` only removes adjacent duplicates; sorting by entry time tends to group duplicates reasonably, but it’s not guaranteed if two distinct cells have the same entry time. Still, the candidate-generation strategy typically avoids problematic repeats; that said, this is a mild fragility of the approach.)

### Degenerate segment case
If \(A==B\) (length ~ 0):
- The beetle doesn’t travel; it “passes” a cell only if the starting point lies strictly inside a cell.
- If the point is on a grid line (`x` or `y` integer), it lies on boundary => no cell.
- If `n != 1`, no solution because it can’t pass multiple cells.
- Else output `floor(x1), floor(y1)`.

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/geometry/point.hpp>

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 coord_t = double;

struct Point {
    static constexpr coord_t eps = 1e-9;

    coord_t x, y;
    Point(coord_t x = 0, coord_t y = 0) : x(x), y(y) {}

    Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
    Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
    Point operator*(coord_t c) const { return Point(x * c, y * c); }
    Point operator/(coord_t c) const { return Point(x / c, y / c); }

    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    bool operator==(const Point& p) const { return x == p.x && y == p.y; }
    bool operator!=(const Point& p) const { return x != p.x || y != p.y; }
    bool operator<(const Point& p) const {
        return x != p.x ? x < p.x : y < p.y;
    }
    bool operator>(const Point& p) const {
        return x != p.x ? x > p.x : y > p.y;
    }
    bool operator<=(const Point& p) const {
        return x != p.x ? x < p.x : y <= p.y;
    }
    bool operator>=(const Point& p) const {
        return x != p.x ? x > p.x : y >= p.y;
    }

    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }
    coord_t angle() const { return atan2(y, x); }

    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    Point perp() const { return Point(-y, x); }
    Point unit() const { return *this / norm(); }
    Point normal() const { return perp().unit(); }
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    friend ostream& operator<<(ostream& os, const Point& p) {
        return os << p.x << ' ' << p.y;
    }
    friend istream& operator>>(istream& is, Point& p) {
        return is >> p.x >> p.y;
    }

    friend int ccw(const Point& a, const Point& b, const Point& c) {
        coord_t v = (b - a) ^ (c - a);
        if(-eps <= v && v <= eps) {
            return 0;
        } else if(v > 0) {
            return 1;
        } else {
            return -1;
        }
    }

    friend bool point_on_segment(
        const Point& a, const Point& b, const Point& p
    ) {
        return ccw(a, b, p) == 0 && p.x >= min(a.x, b.x) - eps &&
               p.x <= max(a.x, b.x) + eps && p.y >= min(a.y, b.y) - eps &&
               p.y <= max(a.y, b.y) + eps;
    }

    friend bool point_in_triangle(
        const Point& a, const Point& b, const Point& c, const Point& p
    ) {
        int d1 = ccw(a, b, p);
        int d2 = ccw(b, c, p);
        int d3 = ccw(c, a, p);
        return (d1 >= 0 && d2 >= 0 && d3 >= 0) ||
               (d1 <= 0 && d2 <= 0 && d3 <= 0);
    }

    friend Point line_line_intersection(
        const Point& a1, const Point& b1, const Point& a2, const Point& b2
    ) {
        return a1 +
               (b1 - a1) * ((a2 - a1) ^ (b2 - a2)) / ((b1 - a1) ^ (b2 - a2));
    }

    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    friend Point circumcenter(const Point& a, const Point& b, const Point& c) {
        Point mid_ab = (a + b) / 2.0;
        Point mid_ac = (a + c) / 2.0;
        Point perp_ab = (b - a).perp();
        Point perp_ac = (c - a).perp();
        return line_line_intersection(
            mid_ab, mid_ab + perp_ab, mid_ac, mid_ac + perp_ac
        );
    }
};

int n;
Point a, b;

void read() { cin >> a >> b >> n; }

void solve() {
    // This is a simple problem - we can simply simulate the n steps, by walking
    // with distance 1.0 in the direction to x1,y1. We look at the 9 adjacent
    // cells, and keep a visited map to not over count (or we can just remove
    // the later on). When we get to N points, we will sort them by distance
    // from x1,y1. We have to be careful about precision issues, so we opt for
    // integer computation.

    Point dir = b - a;
    double len = dir.norm();
    constexpr double eps = 1e-9;

    if(len < eps) {
        bool on_boundary =
            abs(a.x - round(a.x)) < eps || abs(a.y - round(a.y)) < eps;
        if(on_boundary || n != 1) {
            cout << "no solution" << endl;
        } else {
            cout << (int)floor(a.x) << " " << (int)floor(a.y) << endl;
        }
        return;
    }

    vector<pair<int, int>> order;
    double dt = 1.0 / len;

    int x1 = (int)a.x, y1 = (int)a.y, x2 = (int)b.x, y2 = (int)b.y;
    int dx = x2 - x1, dy = y2 - y1;

    auto intersects_cell = [&](int ncx, int ncy) -> bool {
        int64_t t_min_num = 0, t_min_den = 1;
        int64_t t_max_num = 1, t_max_den = 1;

        if(dx != 0) {
            int64_t t1_num = ncx - x1, t2_num = ncx + 1 - x1;
            int64_t den = dx;
            if(den < 0) {
                t1_num = -t1_num;
                t2_num = -t2_num;
                den = -den;
            }
            if(t1_num > t2_num) {
                swap(t1_num, t2_num);
            }
            if(t1_num * t_min_den > t_min_num * den) {
                t_min_num = t1_num;
                t_min_den = den;
            }
            if(t2_num * t_max_den < t_max_num * den) {
                t_max_num = t2_num;
                t_max_den = den;
            }
        } else if(x1 <= ncx || x1 >= ncx + 1) {
            return false;
        }

        if(dy != 0) {
            int64_t t1_num = ncy - y1, t2_num = ncy + 1 - y1;
            int64_t den = dy;
            if(den < 0) {
                t1_num = -t1_num;
                t2_num = -t2_num;
                den = -den;
            }
            if(t1_num > t2_num) {
                swap(t1_num, t2_num);
            }
            if(t1_num * t_min_den > t_min_num * den) {
                t_min_num = t1_num;
                t_min_den = den;
            }
            if(t2_num * t_max_den < t_max_num * den) {
                t_max_num = t2_num;
                t_max_den = den;
            }
        } else if(y1 <= ncy || y1 >= ncy + 1) {
            return false;
        }

        return t_min_num * t_max_den < t_max_num * t_min_den;
    };

    auto get_entry_t = [&](int ncx, int ncy) -> double {
        double t_min = 0;
        if(dx != 0) {
            double t1 = (double)(ncx - x1) / dx;
            double t2 = (double)(ncx + 1 - x1) / dx;
            t_min = max(t_min, min(t1, t2));
        }
        if(dy != 0) {
            double t1 = (double)(ncy - y1) / dy;
            double t2 = (double)(ncy + 1 - y1) / dy;
            t_min = max(t_min, min(t1, t2));
        }
        return t_min;
    };

    for(double t = 0; t <= 1.0 + eps && (int)order.size() < 8 * n; t += dt) {
        double x = a.x + dir.x * t;
        double y = a.y + dir.y * t;
        int cx = (int)floor(x);
        int cy = (int)floor(y);

        for(int dcx = -1; dcx <= 1; dcx++) {
            for(int dcy = -1; dcy <= 1; dcy++) {
                int ncx = cx + dcx;
                int ncy = cy + dcy;
                if(intersects_cell(ncx, ncy)) {
                    order.push_back({ncx, ncy});
                }
            }
        }
    }

    sort(order.begin(), order.end(), [&](const auto& a, const auto& b) {
        return get_entry_t(a.first, a.second) < get_entry_t(b.first, b.second);
    });
    order.erase(unique(order.begin(), order.end()), order.end());

    if((int)order.size() < n) {
        cout << "no solution" << endl;
    } else {
        cout << order[n - 1].first << " " << order[n - 1].second << 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 (solves the problem) with detailed comments

This Python version uses a **standard exact grid-walk (2D DDA)** that is simpler and more robust: it advances from one crossed cell to the next by jumping to the next vertical or horizontal grid line intersection. It properly handles “do not count borders” by slightly biasing the start point if it lies exactly on a grid line.

```python
import sys
import math

def solve():
    x1, y1, x2, y2, n = map(int, sys.stdin.read().split())

    # Degenerate segment: no movement
    if x1 == x2 and y1 == y2:
        # If the point is on any grid line, it's on boundary -> no cell visited
        if x1 == int(x1) or y1 == int(y1):  # always true for ints, but keep logic explicit
            print("no solution")
            return
        # For non-integer inputs we'd return floor; with integer inputs it's always boundary
        print("no solution")
        return

    dx = x2 - x1
    dy = y2 - y1

    # We traverse cells along the segment from (x1,y1) to (x2,y2).
    # Cells are open interiors; borders don't count.
    #
    # To avoid ambiguous cases when starting exactly on a grid line (always true here since ints),
    # we conceptually start from a tiny epsilon inside the segment direction.
    # That ensures we pick the first interior cell consistently.
    eps = 1e-12
    sx = x1 + eps * dx
    sy = y1 + eps * dy

    # Current cell indices
    cx = math.floor(sx)
    cy = math.floor(sy)

    # Determine stepping direction in grid
    step_x = 1 if dx > 0 else -1
    step_y = 1 if dy > 0 else -1

    # t in [0,1] parameterization; we compare next crossings in this parameter space.
    # For DDA we need:
    # - t_max_x: t at which we cross the next vertical grid line
    # - t_max_y: t at which we cross the next horizontal grid line
    # - t_delta_x: how much t increases between successive vertical crossings
    # - t_delta_y: how much t increases between successive horizontal crossings
    #
    # Segment: x(t)=x1+dx*t, so crossing x = k happens at t=(k-x1)/dx.
    if dx != 0:
        # Next vertical grid line after starting point in direction step_x
        next_vline = (cx + 1) if step_x > 0 else cx
        t_max_x = (next_vline - sx) / dx  # dx has sign
        t_delta_x = 1.0 / abs(dx)         # parameter increase per one cell in x
    else:
        t_max_x = float('inf')
        t_delta_x = float('inf')

    if dy != 0:
        next_hline = (cy + 1) if step_y > 0 else cy
        t_max_y = (next_hline - sy) / dy
        t_delta_y = 1.0 / abs(dy)
    else:
        t_max_y = float('inf')
        t_delta_y = float('inf')

    # Collect visited cells in order until we reach n or finish the segment.
    visited = 0

    # We also need to stop when t passes 1 (end point). Since we started at epsilon,
    # we stop when the next boundary crossing would happen after t>1.
    t = 0.0
    while True:
        # The current (cx,cy) is the cell containing the current interior point.
        visited += 1
        if visited == n:
            print(cx, cy)
            return

        # Decide which grid line we cross next:
        # - If t_max_x < t_max_y: cross vertical line first => move in x
        # - If t_max_y < t_max_x: cross horizontal line first => move in y
        # - If equal: cross a corner (both) => move in both
        if t_max_x < t_max_y:
            t = t_max_x
            if t >= 1.0 - 1e-15:
                break
            cx += step_x
            t_max_x += t_delta_x
        elif t_max_y < t_max_x:
            t = t_max_y
            if t >= 1.0 - 1e-15:
                break
            cy += step_y
            t_max_y += t_delta_y
        else:
            # Crossing exactly at a corner: the segment touches a grid corner.
            # Borders/corners aren't cells, but immediately after crossing we enter
            # the diagonally adjacent cell, so we advance both coordinates.
            t = t_max_x
            if t >= 1.0 - 1e-15:
                break
            cx += step_x
            cy += step_y
            t_max_x += t_delta_x
            t_max_y += t_delta_y

    print("no solution")

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

---

## 5) Compressed editorial

Model each cell as open unit square \((c_x,c_x+1)\times(c_y,c_y+1)\). The segment \(A\to B\) “passes” a cell iff there exists \(t\in(0,1)\) with both coordinates strictly inside that square.

The provided solution:
- Samples points along the segment at about one per unit length (`dt=1/|AB|`).
- For each sample, considers the 3×3 neighborhood of the sample’s cell as candidates.
- Filters candidates with an exact slab-intersection test using integer/rational comparisons, enforcing strict interior (`t_min < t_max`).
- Sorts remaining cells by entry time `t_entry` (earliest \(t\) entering the cell’s box), uniques, and outputs the \(n\)-th; otherwise prints `no solution`.
- Handles degenerate \(A=B\): only one cell if point lies strictly inside, else no solution.