Code

#include <atcoder/modint>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;

using mint = modint998244353;

// Tip: Instead of dividing by 2 everytime, we precompute the inverse, and
// multiply by it. In other words, we directly multiply by 1/2, since this is
// a constant and will be used in multiple places.
const mint p = mint(1) / 2;

void solve(int n) {
    // dp[l][r] is the answer for the person for whom l people are standing at
    // the front and r people are standing at the back.
    vector<vector<mint>> dp(n + 1, vector<mint>(n + 1, 0));

    dp[0][0] = 1;

    // Solve on each anti-diagonal.
    // We are solving in increasing order of k, where k = l + r.
    for (int k = 1; k <= n; k++) {
        // Assume that dp[0][k] = x
        // At each step, we make sure that the answer is Ax + B.
        mint A = 1, B = 0;

        int i = 1, j = k - 1;
        while (i <= n && j >= 0) {
            // dp[i][j] = dp[i - 1][j]/2 + dp[i - 1][j + 1]/2;
            // The answer for dp[i - 1][j] is already computed, we can use it.
            A = A * p;
            B = (B + dp[i - 1][j]) * p;

            i++;
            j--;
        }

        // 2*x = A*x + B
        // x = B/(2 - A)
        dp[0][k] = B / (mint(2) - A);

        // Now, backtrack and populate all values.
        i = 1, j = k - 1;
        while (i <= n && j >= 0) {
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) * p;

            i++;
            j--;
        }
    }

    int front = 0, back = n - 1;
    for (int i = 1; i <= n; i++) {
        cout << dp[front][back].val() << " ";
        front++;
        back--;
    }
    cout << "\n";
}

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

    int n;
    cin >> n;

    solve(n);
    return 0;
}

#include <atcoder/modint>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;

using mint = modint998244353;

// Tip: Instead of dividing by 2 everytime, we precompute the inverse, and
// multiply by it. In other words, we directly multiply by 1/2, since this is
// a constant and will be used in multiple places.
const mint p = mint(1) / 2;

void solve(int n) {
    // dp[l][r] is the answer for the person for whom l people are standing at
    // the front and r people are standing at the back.
    vector<vector<mint>> dp(n + 1, vector<mint>(n + 1, 0));

    dp[0][0] = 1;

    // Solve on each anti-diagonal.
    // We are solving in increasing order of k, where k = l + r.
    for (int k = 1; k <= n; k++) {
        // Assume that dp[0][k] = x
        // At each step, we make sure that the answer is Ax + B.
        mint A = 1, B = 0;

        int i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            // dp[i][j] = dp[i - 1][j]/2 + dp[i - 1][j + 1]/2;
            // The answer for dp[i - 1][j] is already computed, we can use it.
            A = A * p;
            B = (B + dp[i - 1][j]) * p;

            i++;
            j--;
        }

        // 2*x = A*x + B
        // x = B/(2 - A)
        dp[0][k] = B / (mint(2) - A);

        // Now, backtrack and populate all values.
        i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) * p;

            i++;
            j--;
        }
    }

    // Now, all values in  the upper triangle is populated correctly.
    // Let's fix the lower triangle now.
    // This isn't really needed, as we don't use the lower triangle.
    // But better safe than sorry.
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) * p;
        }
    }

    int front = 0, back = n - 1;
    for (int i = 1; i <= n; i++) {
        cout << dp[front][back].val() << " ";
        front++;
        back--;
    }
    cout << "\n";
}

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

    int n;
    cin >> n;

    solve(n);
    return 0;
}

#include <atcoder/modint>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;

// Tip: Avoid modular divison as much as possible.
using mint = modint998244353;

const mint p = mint(2);

void solve(int n) {
    // dp[l][r] is the answer for the person for whom l people are standing at
    // the front and r people are standing at the back.
    vector<vector<mint>> dp(n + 1, vector<mint>(n + 1, 0));

    dp[0][0] = 1;

    // Solve on each anti-diagonal.
    // We are solving in increasing order of k, where k = l + r.
    for (int k = 1; k <= n; k++) {
        // Assume that dp[0][k] = x
        // At each step, we make sure that the answer is Ax + B.
        mint A = 1, B = 0;

        int i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            // dp[i][j] = dp[i - 1][j]/2 + dp[i - 1][j + 1]/2;
            // The answer for dp[i - 1][j] is already computed, we can use it.
            // Combining 2 divisons into one to save time.
            A = A / p;
            B = (B + dp[i - 1][j]) / p;

            i++;
            j--;
        }

        // 2*x = A*x + B
        // x = B/(2 - A)
        dp[0][k] = B / (mint(2) - A);

        // Now, backtrack and populate all values.
        i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            // Combining 2 divisons into one to save time.
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) / p;

            i++;
            j--;
        }
    }

    // Now, all values in  the upper triangle is populated correctly.
    // Let's fix the lower triangle now.
    // This isn't really needed, as we don't need the lower triangle.
    // But adding this results in TLE.

    // for (int i = 1; i <= n; i++) {
    //     for (int j = 0; j <= n; j++) {
    //         dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) / p;
    //     }
    // }

    int front = 0, back = n - 1;
    for (int i = 1; i <= n; i++) {
        cout << dp[front][back].val() << " ";
        front++;
        back--;
    }
    cout << "\n";
}

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

    int n;
    cin >> n;

    solve(n);
    return 0;
}

#include <atcoder/modint>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;

using mint = modint998244353;

const mint p = mint(2);

void solve(int n) {
    // dp[l][r] is the answer for the person for whom l people are standing at
    // the front and r people are standing at the back.
    vector<vector<mint>> dp(n + 1, vector<mint>(n + 1, 0));

    dp[0][0] = 1;

    // Solve on each anti-diagonal.
    // We are solving in increasing order of k, where k = l + r.
    for (int k = 1; k <= n; k++) {
        // Assume that dp[0][k] = x
        // At each step, we make sure that the answer is Ax + B.
        mint A = 1, B = 0;

        int i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            // dp[i][j] = p*dp[i - 1][j] + p*dp[i - 1][j + 1]
            // The answer for dp[i - 1][j] is already computed, we can use it.
            A = A / p;
            B = (B + dp[i - 1][j]) / p;

            i++;
            j--;
        }

        // 2*x = A*x + B
        // x = B/(2 - A)
        dp[0][k] = B / (mint(2) - A);

        // Now, backtrack and populate all values.
        i = 0, j = k;
        i++;
        j--;
        while (i <= n && j >= 0) {
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) / p;

            i++;
            j--;
        }
    }

    // Now, all values in  the upper triangle is populated correctly.
    // Let's fix the lower triangle now.
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) / p;
        }
    }

    int front = 0, back = n - 1;
    for (int i = 1; i <= n; i++) {
        cout << dp[front][back].val() << " ";
        front++;
        back--;
    }
    cout << "\n";
}

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

    int n;
    cin >> n;

    solve(n);
    return 0;
}