## 1. Abridged problem statement

Given a `q × c` rectangular field, find the maximum number of `n × 1` rectangular graves that can be placed inside it. Graves must be axis-parallel, so each grave is either horizontal or vertical.

Input contains exactly three integers `q`, `c`, `n`, each possibly extremely large:

- `0 ≤ q, c ≤ 10^1000`
- `1 ≤ n ≤ 10^1000`

Output the maximum number of graves.

---

## 2. Detailed editorial

We need maximize the number of axis-parallel `n × 1` strips packed into a `q × c` rectangle.

Because the numbers have up to `1000` decimal digits, the main challenge is not complexity but deriving a formula that only needs big integer arithmetic.

---

### Case 1: One side is smaller than `n`

If `min(q, c) < n`, then a grave cannot be placed along the short side.

So every grave must be parallel to the longer side.

If the rectangle has dimensions:

```text
short × long
```

then each of the `short` rows/columns can contain:

```text
floor(long / n)
```

graves.

Therefore:

```text
answer = min(q, c) * floor(max(q, c) / n)
```

This also correctly gives `0` if both sides are smaller than `n`.

---

### Case 2: Both sides are at least `n`

Assume:

```text
q ≥ n and c ≥ n
```

Write:

```text
q = x * n + s
c = y * n + t
```

where:

```text
s = q mod n
t = c mod n
0 ≤ s, t < n
```

The answer is:

```text
(q * c - s * t) / n + max(0, s + t - n)
```

Equivalently, since each grave covers `n` cells:

```text
answer = (q * c - deficiency) / n
```

where:

```text
deficiency = min(s * t, (n - s) * (n - t))
```

---

### Why is this an upper bound?

Color the infinite grid using `n` colors:

```text
color(i, j) = (i + j) mod n
```

Any horizontal or vertical `n × 1` grave covers `n` consecutive cells, so it contains exactly one cell of each color.

Therefore, if there are `T` graves, then for every color `k`:

```text
T ≤ number of cells of color k inside the q × c rectangle
```

So:

```text
T ≤ minimum color-class size
```

Now count the number of cells of each color.

Let:

```text
q = x * n + s
c = y * n + t
```

Rows of each residue modulo `n` appear either `x` or `x + 1` times. Exactly `s` row residues appear `x + 1` times.

Similarly, columns of each residue appear either `y` or `y + 1` times. Exactly `t` column residues appear `y + 1` times.

For any color `k`, its count has a common base part:

```text
n * x * y + x * t + y * s
```

plus an extra term `Q_k`, which counts overlaps between two cyclic intervals of lengths `s` and `t`.

The minimum possible overlap of two cyclic arcs of lengths `s` and `t` on a cycle of length `n` is:

```text
max(0, s + t - n)
```

Therefore:

```text
T ≤ n*x*y + x*t + y*s + max(0, s + t - n)
```

This simplifies to:

```text
T ≤ (q * c - s * t) / n + max(0, s + t - n)
```

---

### Why is this bound achievable?

When both `q ≥ n` and `c ≥ n`, we can construct a packing matching the bound.

First tile the large full multiples of `n`.

Write:

```text
q = x*n + s
c = y*n + t
```

Because `x, y ≥ 1`, remove and completely tile:

1. The top `(x - 1) * n` rows using vertical graves.
2. From the remaining rectangle, the left `(y - 1) * n` columns using horizontal graves.

This leaves only a smaller corner rectangle:

```text
(n + s) × (n + t)
```

Now we need optimally fill this corner.

There are two natural fillings.

---

#### Simple filling

Fill:

- one horizontal grave in each row, covering `n` cells;
- one vertical grave in each remaining extra column.

This leaves an empty `s × t` corner.

Deficiency:

```text
s * t
```

---

#### Pinwheel filling

If `s + t > n`, we can do better by arranging four arms in a pinwheel pattern.

The empty hole then has size:

```text
(n - s) × (n - t)
```

Deficiency:

```text
(n - s) * (n - t)
```

This is better exactly when:

```text
(n - s) * (n - t) < s * t
```

which simplifies to:

```text
s + t > n
```

So the optimal deficiency is:

```text
min(s * t, (n - s) * (n - t))
```

Thus the maximum number of graves is:

```text
(q * c - min(s * t, (n - s) * (n - t))) / n
```

which is the same as:

```text
(q * c - s * t) / n + max(0, s + t - n)
```

---

## 3. C++ Solution

Since the inputs have up to `1000` decimal digits and C++ has no built-in arbitrary-precision type, the solution carries a self-contained `bigint` struct (base-`10^9` limbs with Karatsuba multiplication and long division). The actual algorithm at the bottom is short: it is exactly the case analysis above.

```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;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

bigint q, c, n;

void read() { cin >> q >> c >> n; }

void solve() {
    // We pack 1xn strips into a qxc board, maximizing the count. Color cell
    // (i, j) with (i + j) mod n. A strip covers n consecutive cells in one
    // line, so it hits each of the n colors exactly once. Hence every placed
    // strip uses one cell of color k, and if N_k is the number of board cells
    // of color k, the number of strips T satisfies T <= N_k for all k, i.e.
    //
    //     T <= min_k N_k.
    //
    // Write x = q / n, s = q % n, y = c / n, t = c % n. Rows contribute residue
    // a with multiplicity (x + [a < s]) and columns residue b with (y + [b <
    // t]), so N_k = n*x*y + x*t + y*s + Q_k, where Q_k counts b in [0, t) with
    // (k - b) mod n in [0, s). As k slides, that is the overlap of a length-s
    // cyclic arc with the length-t interval [0, t) in Z_n; two arcs of lengths
    // s and t on a cycle of length n overlap in at least max(0, s + t - n)
    // cells, and that minimum is attained. Therefore
    //
    //     T <= (q*c - s*t) / n + max(0, s + t - n).
    //
    // When both q >= n and c >= n this bound is achievable. Peel off full
    // slabs whose side is a multiple of n -- the top n*(x - 1) rows (vertical
    // strips) and, in the rest, the left n*(y - 1) columns (horizontal strips)
    // -- both tile completely, leaving a single (n + s) x (n + t) corner block.
    // Filling that block trivially (a strip along each row's left part and each
    // column's top part) leaves the s x t sub-corner empty: deficiency s*t.
    // Pinwheeling it instead leaves only an (n - s) x (n - t) hole: deficiency
    // (n - s)*(n - t) = s*t - n*(s + t - n). Each arm has a side of length n,
    // so it tiles completely; the pinwheel wins (by s + t - n strips) exactly
    // when s + t > n. With s = q % n, t = c % n the corner block looks like:
    //
    //          0            t           n          n+t
    //        0 +------------+-----------+-----------+
    //          |                        |           |
    //          |          TOP           |   RIGHT   |
    //          |         s x n          |   n x t   |
    //          |      (horizontal)      | (vertical)|
    //        s +------------+-----------+           |
    //          |            |           |           |
    //          |    LEFT    |   HOLE    |           |
    //          |   n x t    |  (n-s) x  |           |
    //          | (vertical) |   (n-t)   |           |
    //        n |            +-----------+-----------+
    //          |            |                       |
    //          |            |        BOTTOM         |
    //          |            |         s x n         |
    //          |            |     (horizontal)      |
    //      n+s +------------+-----------------------+
    //
    // So the deficiency is min(s*t, (n - s)*(n - t)). This meets the upper
    // bound exactly: the pinwheel saves s*t - (n - s)*(n - t) = n*(s + t - n)
    // cells over the trivial fill, i.e. s + t - n extra strips, which is
    // precisely the min_k Q_k = max(0, s + t - n) bonus times n, so the
    // construction attains min_k N_k and is optimal. (For s = t = 2, n = 3 this
    // is the pinwheel on a 5x5 board with arms 2x3, 2x3, 3x2, 3x2 around a 1x1
    // hole: only the centre empty.)
    //
    // If min(q, c) < n only the orientation along the long side fits, because
    // the short side cannot hold a strip; the corner bonus needs both
    // orientations and vanishes. Each of the min(q, c) lines independently
    // holds floor(max(q, c) / n) strips, giving min(q, c) * floor(max(q, c) /
    // n), which is 0 when both sides are below n.

    bigint ans;
    if(q >= n && c >= n) {
        bigint s = q % n, t = c % n;
        ans = (q * c - s * t) / n;
        if(s + t > n) {
            ans += s + t - n;
        }
    } else {
        ans = min(q, c) * (max(q, c) / n);
    }

    cout << ans << '\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 with detailed comments

Python integers are arbitrary precision by default, so the implementation is very short.

```python
# Read q: one side of the cemetery rectangle.
q = int(input())

# Read c: the other side of the cemetery rectangle.
c = int(input())

# Read n: grave length, each grave has size n x 1.
n = int(input())

# If both sides are at least n, we can use the full optimal formula.
if q >= n and c >= n:
    # s is the leftover part of q after full blocks of length n.
    s = q % n

    # t is the leftover part of c after full blocks of length n.
    t = c % n

    # Basic construction leaves an s * t uncovered corner.
    ans = (q * c - s * t) // n

    # If s + t > n, the pinwheel construction improves the packing.
    if s + t > n:
        ans += s + t - n

# Otherwise, one side is too short to fit a grave across it.
else:
    # The shorter side gives the number of independent rows/columns.
    short = min(q, c)

    # The longer side is where graves are placed.
    long = max(q, c)

    # Each of the short lines contains floor(long / n) graves.
    ans = short * (long // n)

# Output the maximum possible number of graves.
print(ans)
```

---

## 5. Compressed editorial

Let the rectangle be `q × c`, grave size `n × 1`.

If `min(q, c) < n`, only one orientation is possible, so:

```text
answer = min(q, c) * floor(max(q, c) / n)
```

Otherwise, write:

```text
q = x*n + s
c = y*n + t
```

with `0 ≤ s,t < n`.

Color cell `(i,j)` by `(i+j) mod n`. Every grave covers exactly one cell of each color, so the number of graves is at most the smallest color class. Counting color classes gives:

```text
answer ≤ (q*c - s*t)/n + max(0, s+t-n)
```

This bound is achievable. Tile all full `n`-multiple slabs, leaving a corner of size:

```text
(n+s) × (n+t)
```

This corner can be filled leaving either:

```text
s*t
```

cells empty, or, using a pinwheel construction when `s+t > n`:

```text
(n-s)*(n-t)
```

cells empty.

Thus the optimal formula for `q,c ≥ n` is:

```text
answer = (q*c - s*t)/n + max(0, s+t-n)
```

Use arbitrary-precision integers because inputs have up to `1000` digits.
