1. Concise Problem Statement  
Given an n×n chessboard, count the number of ways to place k bishops so that no two attack each other. Bishops attack along diagonals. Output the exact count.

2. Detailed Editorial  

Overview  
Because bishops move only along diagonals, any two bishops on different-color squares can never attack each other. On an n×n board there are two independent sets of diagonals—“even” (say, black-square) diagonals and “odd” (white-square) diagonals. We split the problem into two subproblems: choose i bishops on even diagonals and k−i bishops on odd diagonals, for all i, and multiply the counts, then sum over i.

Counting on one parity  
Label the diagonals of one color by their lengths in increasing order. On an n×n board:

 • For “even” diagonals (starting from the very top corner), their lengths go 1,2,3,…,n−1,n,n−1,…,2,1, but we only take every other diagonal depending on color.  
 • You end up with a list per_diag of m diagonals, each with a certain number of available squares v1, v2, …, vm.  

We wish to place j bishops among the first t diagonals so that no two share a diagonal—this means at most one bishop per diagonal. But bishops on different diagonals of the same color still attack if they share the *other* direction diagonal. However, by construction of our list (one color alone), any two chosen squares automatically lie on distinct diagonals in both directions if we never choose more than one from the same diagonal. Thus it reduces to “choose j diagonals out of t” but weighted by how many squares are on each diagonal.

Dynamic Programming  
Let dp[t][j] = number of ways to choose j bishops among the first t diagonals. Each diagonal t has vt squares:

 Base: dp[0][0] = 1; dp[0][j>0] = 0.  
 Transition: when considering diagonal t with vt squares, you either place no bishop there (dp[t−1][j] ways), or place one bishop there (choose one of vt squares, but note that if you already placed (j−1) bishops elsewhere, vt reduces by (j−1) because no two bishops on the same “anti”-diagonal—but this detail is built into the counting of per-diagonal list). Concretely:
   dp[t][j] = dp[t−1][j]            // skip this diagonal  
            + dp[t−1][j−1] * (vt − (j−1))  
The term (vt − (j−1)) accounts for the fact that j−1 bishops already chosen forbid one square on that diagonal for each of them, but because all chosen bishops are on distinct “other” diagonals, exactly j−1 squares on the vt are attacked, leaving (vt − (j−1)) valid choices.

Combine parities  
Compute dp_even and dp_odd. Then the final answer is  
   sum_{i=0..k} dp_even[m_even][i] * dp_odd[m_odd][k−i].  

Time complexity O(n·k).

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;
};

// 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;
    }
};

int n, k;

void read() { cin >> n >> k; }

vector<bigint> solve_one_parity(int p) {
    vector<int> per_diag;
    for(int i = p; i < n; i += 2) {
        per_diag.push_back(i + 1);
        if(i != n - 1) {
            per_diag.push_back(i + 1);
        }
    }

    vector<bigint> prev(k + 1, bigint(0)), cur(k + 1, bigint(0));
    prev[0] = bigint(1);
    for(int v: per_diag) {
        cur.assign(k + 1, bigint(0));
        cur[0] = bigint(1);
        for(int j = 1; j <= min(k, v); j++) {
            cur[j] = prev[j] + prev[j - 1] * (v - j + 1);
        }

        swap(prev, cur);
    }

    return prev;
}

void solve() {
    // Bishops attack along the two diagonal directions. Rotating the board 45
    // degrees turns the diagonals into independent rows so the board splits by
    // cell colour ((r+c) parity) into two sub-boards whose diagonals never
    // interact. For one colour the available cells form anti-diagonals whose
    // lengths are 1, 2, 2, 3, 3, ... (per_diag): processing these diagonals in
    // increasing length order, dp[j] = number of ways to place j non-attacking
    // bishops, with the transition cur[j] = prev[j] + prev[j-1] * (v - j + 1),
    // since a new diagonal of length v offers (v - j + 1) free squares given
    // j-1 already placed on shorter diagonals. Counts grow factorially so we
    // accumulate them with the vendored bigint. The final answer for k bishops
    // is the convolution sum over i of black[i] * white[k - i].

    if(k > 2 * n - 1) {
        cout << 0 << '\n';
        return;
    }

    vector<bigint> data_0 = solve_one_parity(0);
    vector<bigint> data_1 = solve_one_parity(1);

    bigint ans(0);
    for(int i = 0; i <= k; i++) {
        ans += data_0[i] * data_1[k - i];
    }

    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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
def solve_one_parity(n, k, start):
    """
    Build dp for diagonals of one color.
    start = 0 for “even” diagonals, 1 for “odd”.
    Returns dp table where dp[i][j] = ways to place j bishops
    on the first i diagonals of this parity.
    """
    per_diag = []
    # Collect lengths of diagonals of this parity
    # Diagonal indices run from 0 to 2n-2
    for d in range(start, 2*n - 1, 2):
        if d < n:
            length = d + 1
        else:
            length = 2*n - 1 - d
        per_diag.append(length)

    m = len(per_diag)
    # Initialize dp with zeros
    dp = [[0] * (k+1) for _ in range(m+1)]
    dp[0][0] = 1  # zero bishops in zero diagonals

    # Fill DP table
    for i in range(1, m+1):
        v = per_diag[i-1]  # length of the i-th diagonal
        dp[i][0] = 1       # placing 0 bishops is always 1 way
        # for j>=1, either skip or place one bishop
        for j in range(1, k+1):
            # skip this diagonal
            dp[i][j] = dp[i-1][j]
            # place one bishop: we have v squares but j-1 bishops already placed
            # rule out (j-1) attacked squares => (v - (j-1)) choices if >=1
            if j <= v:
                dp[i][j] += dp[i-1][j-1] * (v - (j-1))
    return dp

def solve(n, k):
    # if k exceeds total diagonals on one color, impossible
    if k > 2*n - 1:
        return 0

    # build dp tables for both parities
    dp0 = solve_one_parity(n, k, 0)
    dp1 = solve_one_parity(n, k, 1)

    ways = 0
    m0 = len(dp0) - 1
    m1 = len(dp1) - 1
    # combine counts: choose i on parity0, k-i on parity1
    for i in range(k+1):
        ways += dp0[m0][i] * dp1[m1][k-i]
    return ways

# Read input and output answer
n, k = map(int, input().split())
print(solve(n, k))
```

5. Compressed Editorial  
- Split board into two independent sets of diagonals by color (even/odd).  
- For each parity, list diagonals’ lengths: per_diag.  
- DP: dp[i][j] = ways to place j bishops in first i diagonals;  
    dp[i][j] = dp[i-1][j] + dp[i-1][j-1]*(v_i − (j−1)).  
- Combine both parities: ∑_{i=0..k} dp_even[i] * dp_odd[k−i].  
- Time O(n·k), exact big-integer arithmetic.