Problem A: Threshold Validation

Statement: Evaluate the summation of two integer inputs. Return the calculated value if it remains within or below ten; otherwise, flag an invalid state.

Approach: Direct arithmetic comparison eliminates the need for complex graph algorithms. Computing the aggregate and applying a single conditional branch yields the correct output in constant time.

#include <iostream>

void evaluate_threshold() {
    int val_a, val_b;
    if (!(std::cin >> val_a >> val_b)) return;
    
    int aggregate = val_a + val_b;
    if (aggregate > 10) {
        std::cout << "error\n";
    } else {
        std::cout << aggregate << "\n";
    }
}

Time Complexity: $O(1)$

Problem B: Character Uniqueness Verification

Statement: Analyze a sequance of lowercase Latin letters to determine whether every symbol appears exactly once.

Approach: Leveraging the pigeonhole principle, any string exceeding twenty-six characters must contain repetitions. For shorter sequences, a fixed-size frequency tracker validates uniqueness in linear time without requiring sorting operations. This approach reduces overhead and avoids $O(N \log N)$ comparisons.

#include <iostream>
#include <vector>

bool verify_uniqueness() {
    std::string sequence;
    std::cin >> sequence;
    
    auto size = sequence.length();
    if (size > 26) return false;
    
    std::vector<int> occurrence_map(26, 0);
    for (char ch : sequence) {
        int idx = ch - 'a';
        occurrence_map[idx]++;
        if (occurrence_map[idx] > 1) return false;
    }
    return true;
}

int main() {
    std::cout << (verify_uniqueness() ? "yes\n" : "no\n");
    return 0;
}

Time Complexity: $O(|S|)$

Problem C: Maximizing Non-Multiples

Statement: Given a array of $N$ integers, identify the largest possible subset sum that is strictly not divisible by ten.

Approach: Begin by calculating the totall sum of all provided elements. If this cumulative value modulo ten is non-zero, it immediately satisfies the condition and serves as the optimal answer. When the total is evenly divisible by ten, removing the smallest element that breaks the divisibility pattern will yield the maximum valid remainder. If no such element exists (i.e., all numbers are multiples of ten), the target configuration is unreachable, defaulting the result to zero.

#include <iostream>
#include <limits>

void solve_max_sum() {
    int count;
    std::cin >> count;
    
    long long current_total = 0;
    int min_non_divisor = std::numeric_limits<int>::max();
    bool found_candidate = false;
    
    for (int i = 0; i < count; ++i) {
        int val;
        std::cin >> val;
        current_total += val;
        
        if (val % 10 != 0) {
            if (val < min_non_divisor) {
                min_non_divisor = val;
            }
            found_candidate = true;
        }
    }
    
    if (current_total % 10 != 0) {
        std::cout << current_total << "\n";
    } else if (found_candidate) {
        std::cout << current_total - min_non_divisor << "\n";
    } else {
        std::cout << 0 << "\n";
    }
}

Time Complexity: $O(N)$

Problem D: Optimized Strike Allocation

Statement: Engage $N$ enemies with varying vitality points. Each activation of a special ability deals damage $A$ to a targeted foe and damage $B$ to all other active foes. Calculate the minimum activation count required to reduce every enemy's vitality to zero or below.

Approach: Binary search over the potential activation count $X$. Assume $X$ activations occur globally. Consequently, every enemy sustains at least $X \times B$ damage from area effects. Surviving vitality must be eliminated through precision hits, each dealing net bonus damage of $A - B$. The required precision strikes for enemy $i$ equals $\lceil (h_i - X \times B) / (A - B) \rceil$. Sum these requirements across all enemies. If the total precision count stays within the initial limit $X$, the hypothesis is feasible.

Note: When $A \leq B$, precision strikes cannot compensate for area damage deficits. In this scenario, feasibility strictly depends on whether $X \times B$ alone exceeds all vitality thresholds.

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

bool can_eliminate_all(const std::vector<long long>& hp, long long base_dmg, long long precision_bonus, int limit) {
    long long needed_precision = 0;
    for (long long h : hp) {
        long long remaining = h - static_cast<long long>(limit) * base_dmg;
        if (remaining > 0) {
            needed_precision += (remaining + precision_bonus - 1) / precision_bonus;
            if (needed_precision > limit) return false;
        }
    }
    return needed_precision <= limit;
}

void optimize_strike_count() {
    int N;
    long long A, B;
    std::cin >> N >> A >> B;
    
    std::vector<long long> hp(N);
    for (int i = 0; i < N; ++i) std::cin >> hp[i];
    
    if (A <= B) {
        long long max_hp = 0;
        for (long long h : hp) max_hp = std::max(max_hp, h);
        if (max_hp == 0) {
            std::cout << 0 << "\n";
            return;
        }
        long long res = (max_hp + B - 1) / B;
        std::cout << res << "\n";
        return;
    }
    
    long long low = 0, high = 1e9 + 7;
    long long ans = high;
    
    while (low <= high) {
        long long mid = low + (high - low) / 2;
        if (can_eliminate_all(hp, B, A - B, mid)) {
            ans = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    
    std::cout << ans << "\n";
}

Time Complexity: $O(N \log (\max(h_i)))$