A: Basic Arithmetic and Output

Given integers a, b, c, d, compute (a + b) * (c - d) and output the result followed by the string "Takahashi".

int a = input(), b = input(), c = input(), d = input();
cout << (a + b) * (c - d) << endl;
cout << "Takahashi" << endl;

Time complexity: O(1)

B: Finding Corners in a Grid

Given a grid of characters, locate the top-left and bottom-right corners of a rectangular region where '#' marks the boundaries.

char grid[MAX][MAX];
int top_x = -1, top_y = -1, bot_x = -1, bot_y = -1;

for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
        if (grid[i][j] == '#') {
            if (grid[i-1][j] != '#' && grid[i][j-1] != '#' && grid[i-1][j-1] != '#') {
                top_x = i; top_y = j;
            }
            if (grid[i+1][j] != '#' && grid[i][j+1] != '#' && grid[i+1][j+1] != '#') {
                bot_x = i; bot_y = j;
            }
        }
    }
}
cout << top_x << " " << top_y << endl << bot_x << " " << bot_y << endl;

Time complexity: O(n²)

C: Subset Generation via Bitmasking

Given a number n, generate all x such that x OR n equals n (i.e., all subset of the set bits in n).

long long n = input();
vector<long long=""> subsets;
for (long long i = n; i; i = (i - 1) & n) {
    subsets.push_back(i);
}
subsets.push_back(0);
reverse(subsets.begin(), subsets.end());
for (auto x : subsets) cout << x << endl;
</long>

Time complexity: O(2^(popcount(n)))

D: Connected Components in a Hexagonal Grid

Given a grid where '#' represents nodes, count the number of connected components considering hexagonal adjacency (six directions).

int grid[5005][5005], component_count = 0;
int dx[] = {-1, -1, 0, 0, 1, 1};
int dy[] = {-1, 0, -1, 1, 0, 1};

void dfs(int x, int y) {
    if (grid[x][y] != 1) return;
    grid[x][y] = 2;
    for (int k = 0; k < 6; k++) {
        dfs(x + dx[k], y + dy[k]);
    }
}

int main() {
    // Read input and mark positions
    for (int i = 0; i < n; i++) {
        int x = input() + 1010, y = input() + 1010;
        grid[x][y] = 1;
    }
    for (int i = 1; i < 2050; i++) {
        for (int j = 1; j < 2050; j++) {
            if (grid[i][j] == 1) {
                component_count++;
                dfs(i, j);
            }
        }
    }
    cout << component_count << endl;
}

Time complexity: O(n²)

E: Binary Search for Unique Row and Column

Given a matrix with exactly one row and one column that do not sum to 1, locate these indices using binary search with range queries.

int n = input();
int low_r = 1, high_r = n, mid_r;
while (low_r < high_r) {
    mid_r = (low_r + high_r) / 2;
    cout << "? " << low_r << " " << mid_r << " " << 1 << " " << n << endl;
    fflush(stdout);
    int response = input();
    if (response != (mid_r - low_r + 1)) {
        high_r = mid_r;
    } else {
        low_r = mid_r + 1;
    }
}
int row_ans = low_r;

int low_c = 1, high_c = n, mid_c;
while (low_c < high_c) {
    mid_c = (low_c + high_c) / 2;
    cout << "? " << 1 << " " << n << " " << low_c << " " << mid_c << endl;
    fflush(stdout);
    int response = input();
    if (response != (mid_c - low_c + 1)) {
        high_c = mid_c;
    } else {
        low_c = mid_c + 1;
    }
}
int col_ans = low_c;

cout << "! " << row_ans << " " << col_ans << endl;
fflush(stdout);

Query complexity: O(log n)

F: Summation Over a Grid with Alternating Pattern

Compute the sum of grid values in a rectangle where each element (i-1)*m + j is included only if (i+j-1) is odd.

const int MOD = 998244353;
long long n, m, q;

long long compute_range(long long x, long long y) {
    if (x <= 0 || y <= 0) return 0;
    long long full_rows = x / 2;
    long long base = ((full_rows * m) % MOD + (full_rows * (full_rows + 1)) % MOD) % MOD;
    long long step = (2 * m * full_rows) % MOD;
    long long sum_full = (base + (base + step * (full_rows - 1)) % MOD * full_rows % MOD * inv(2, MOD) % MOD;
    if (x % 2 == 1) {
        long long last_row = (x / 2) * (x / 2 - 1) % MOD;
        last_row = (last_row + (x / 2) * ((y - 1) * m + 1) % MOD) % MOD;
        sum_full = (sum_full + last_row) % MOD;
    }
    return sum_full;
}

int main() {
    cin >> n >> m >> q;
    while (q--) {
        long long a, b, c, d;
        cin >> a >> b >> c >> d;
        long long result = (compute_range(b, d) - compute_range(a-1, d) - compute_range(b, c-1) + compute_range(a-1, c-1)) % MOD;
        cout << (result + MOD) % MOD << endl;
    }
}

Time complexity: O(q)

G: Dynamic Programming with Limited Value Range

For each k from 0 to m, choose between A_i and B_i for each element to achieve sum k with minimal selections from B, given that sum(A) + sum(B) = m.

Let diff_i = B_i - A_i and base = sum(A). Use dynamic programming where state f[i][x] represents the minimal changes to achieve sum x after i elements.

Key observations: The total absolute differences sum to at most m, and the number of distinct differences is O(√m).

// Conceptual code outline - full implementation depends on input constraints
vector<int> dp(m+1, INF);
dp[base] = 0;
for (int i = 0; i < n; i++) {
    vector<int> next_dp = dp;
    for (int j = 0; j <= m; j++) {
        if (dp[j] != INF && j + diff[i] >= 0 && j + diff[i] <= m) {
            next_dp[j + diff[i]] = min(next_dp[j + diff[i]], dp[j] + 1);
        }
    }
    dp = move(next_dp);
}
for (int k = 0; k <= m; k++) {
    cout << (dp[k] == INF ? -1 : dp[k]) << endl;
}
</int></int>

Time complexity: O(n * m)