A: Minimum Increments to Surpass the Suffix Peak

The task requires determining how many unit additions must be applied to the first element so it strictly exceeds every subsequent value in the sequence. By scanning the subarray starting from the second index, we locate its highest value. The operation count is derived from the gap between that peak and the initial element, ensuring the result never drops below zero.

#include <iostream>
#include <vector>
#include <algorithm>

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int n;
    std::cin >> n;

    std::vector<long long> values(n);
    for (int i = 0; i < n; ++i) {
        std::cin >> values[i];
    }

    long long peak_val = -1;
    for (size_t i = 1; i < values.size(); ++i) {
        peak_val = std::max(peak_val, values[i]);
    }

    long long steps_needed = std::max(0LL, peak_val - values[0] + 1);
    std::cout << steps_needed << '\n';

    return 0;
}

B: Single Winner Determination

This problem simulates a pairwise elimination bracket where each comparison definitively identifies a losing participant. Instead of tracking wins, we maintain a dynamic pool of surviving candidates. Iterating through all reported matchups, we simply discard the designated loser. Once all interactions are processed, a surviving set containing exactly one member indicates a clear victor; any other cardinality implies an unresolved outcome.

#include <iostream>
#include <set>

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int total_contestants, matches_processed;
    if (!(std::cin >> total_contestants >> matches_processed)) return 0;

    std::set<int> active_pool;
    for (int id = 1; id <= total_contestants; ++id) {
        active_pool.insert(id);
    }

    for (int i = 0; i < matches_processed; ++i) {
        int victor, defeated;
        std::cin >> victor >> defeated;
        active_pool.erase(defeated);
    }

    if (active_pool.size() == 1) {
        std::cout << *active_pool.begin() << '\n';
    } else {
        std::cout << "-1\n";
    }

    return 0;
}

C: Near-Uniform Value Distribution

Achieving near-equality across all positions involves targeting the integer quotient of the total sum divided by the array length. Values falling short of this basleine dictate mandatory increase operations. Those already positioned above or at the threshold only require reduction until they rest just above the floor, preventing unnecessary downward moves. When the modulo remainder indicates extra units must be distributed, we verify whether existing high-value elements absorb them or if additional operations are required.

#include <iostream>
#include <vector>

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int size_n;
    std::cin >> size_n;

    std::vector<long long> dataset(size_n);
    long long cumulative_total = 0;
    for (int i = 0; i < size_n; ++i) {
        std::cin >> dataset[i];
        cumulative_total += dataset[i];
    }

    long long target_level = cumulative_total / size_n;
    long long residual_units = cumulative_total % size_n;
    long long ops_count = 0;
    int elevated_elements = 0;

    for (long long val : dataset) {
        if (val < target_level) {
            ops_count += (target_level - val);
        } else {
            elevated_elements++;
        }
    }

    long long adjustment_gap = std::max(0LL, residual_units - elevated_elements);
    ops_count += adjustment_gap;

    std::cout << ops_count << '\n';
    return 0;
}

D: Bitwise Parity Reconstruction

When an interactive system reveals the parity of summations over selected segments of a binary array, the exclusive OR operation serves as the optimal inversion mechanism. Because parity alignment matches XOR accumulation for zero-and-one sequences, strategically omitting individual indices from overlapping windows of width K enables isolated bit extraction. The first K+1 components are resolved by aggregating complementary query results. For trailing positions, a sliding reverse-window technique leverages previously decoded values to nullify known terms and expose the unknown bit.

#include <iostream>
#include <vector>

void submit_query(const std::vector<int>& positions) {
    std::cout << '?';
    for (int p : positions) std::cout << ' ' << p;
    std::cout << '\n' << std::flush;
}

int main() {
    int n, window_k;
    std::cin >> n >> window_k;

    std::vector<int> reconstructed(n, 0);
    std::vector<int> window_indices(window_k);

    // Step 1: Isolate initial k+1 bits via overlapping exclusions
    for (int skip_pos = 0; skip_pos <= window_k; ++skip_pos) {
        window_indices.clear();
        for (int curr = 0; curr <= window_k; ++curr) {
            if (curr != skip_pos) {
                window_indices.push_back(curr + 1);
            }
        }
        submit_query(window_indices);

        int query_outcome;
        std::cin >> query_outcome;
        reconstructed[skip_pos] ^= query_outcome;
    }

    // Step 2: Propagate recovery using sliding backward windows
    for (int frontier = window_k + 1; frontier < n; ++frontier) {
        window_indices.clear();
        for (int off = 0; off < window_k; ++off) {
            window_indices.push_back((frontier - off) + 1);
        }
        submit_query(window_indices);

        std::cin >> reconstructed[frontier];
        for (int off = 1; off < window_k; ++off) {
            reconstructed[frontier] ^= reconstructed[frontier - off];
        }
    }

    std::cout << '!';
    for (int bit : reconstructed) std::cout << ' ' << bit;
    std::cout << '\n' << std::flush;

    return 0;
}