Problem Summary & Key Intuition:

• Summary: Given an operational year Y and a calibration interval k, find the number of years until the next strict future multiple of k.
• Key Intuition: The next strict multiple is given by k — (Y % k).

Detailed Explanation & Complexity:

• Formula: Writing Y = q * k + r with 0 <= r < k, the distance is k — r.
• Time Complexity: O(1) per testcase.
• Space Complexity: O(1) extra space.

C++ Implementation:

 #include <iostream>

using namespace std;

void solve() {
    long long year, interval;
    cin >> year >> interval;
    long long remainder = year % interval;
    long long answer = interval — remainder;
    cout << answer << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int test_cases;
    cin >> test_cases;
    while (test_cases--) {
        solve();
    }
    return 0;
}

Problem Summary & Key Intuition:

• Summary: Given N stone weights and a container weight capacity K, maximize the total count of stones selected such that their total weight does not exceed K.
• Key Intuition: To maximize the item count under a fixed capacity limit, always pick the lightest available items first (classic greedy strategy).

Detailed Explanation & Complexity:

• Detailed Explanation: Sort the array of weights in ascending order w1 <= w2 <= ... <= wN. Greedily accumulate weights starting from the smallest until adding the next stone exceeds capacity K.
• Time Complexity: O(N log N) per testcase due to sorting.
• Space Complexity: O(N) space.

C++ Implementation:

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

using namespace std;

void solve() {
    int n;
    long long capacity;
    cin >> n >> capacity;
    
    vector<long long> weights(n);
    for (int i = 0; i < n; i++) {
        cin >> weights[i];
    }
    
    sort(weights.begin(), weights.end());
    
    long long current_sum = 0;
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        if (current_sum + weights[i] > capacity) {
            break;
        }
        current_sum += weights[i];
        count++;
    }
    
    cout << count << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int test_cases;
    cin >> test_cases;
    while (test_cases--) {
        solve();
    }
    return 0;
}

Problem Summary & Key Intuition:

• Summary: We may flip signs of a prefix of length p and a disjoint suffix of length q (p + q <= n) of an array. Flipping position i costs 2|a_i| if i is even and |a_i| if i is odd. Find the minimum total cost so that the array sums to exactly S, or output -1.
• Key Intuition: Flipping a_i changes the total sum by -2a_i. If total sum is T, the sum of flipped elements must equal K = (T — S) / 2. If T — S is odd, no solution exists.

Detailed Explanation & Complexity:

• Detailed Explanation: Precompute prefix sums/costs and suffix sums/costs. As prefix length p decreases from n down to 0, maintain a hash map storing the minimum cost to achieve each suffix sum seen so far. For each prefix p, query the map for the required remaining sum K — prefSum[p].
• Time Complexity: O(N) amortized per test case.
• Space Complexity: O(N) extra space.

C++ Implementation:

 #include <iostream>
 #include <vector>
 #include <cmath>
 #include <unordered_map>

using namespace std;

const long long INF = 1e18;

void solve() {
    int n;
    long long target_sum;
    cin >> n >> target_sum;
    
    vector<long long> a(n + 1);
    vector<long long> flip_cost(n + 1);
    long long total_sum = 0;
    
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        total_sum += a[i];
        long long val = abs(a[i]);
        flip_cost[i] = (i % 2 == 0) ? 2 * val : val;
    }
    
    long long diff = total_sum — target_sum;
    if (diff % 2 != 0) {
        cout << -1 << "\n";
        return;
    }
    
    long long required_sum = diff / 2;
    
    vector<long long> pref_sum(n + 1, 0), pref_cost(n + 1, 0);
    for (int i = 1; i <= n; i++) {
        pref_sum[i] = pref_sum[i — 1] + a[i];
        pref_cost[i] = pref_cost[i — 1] + flip_cost[i];
    }
    
    vector<long long> suff_sum(n + 2, 0), suff_cost(n + 2, 0);
    for (int i = n; i >= 1; i--) {
        suff_sum[i] = suff_sum[i + 1] + a[i];
        suff_cost[i] = suff_cost[i + 1] + flip_cost[i];
    }
    
    unordered_map<long long, long long> best_suffix_cost;
    long long min_total_cost = INF;
    
    for (int p = n; p >= 0; p--) {
        int idx = p + 1;
        if (idx <= n + 1) {
            long long s_val = suff_sum[idx];
            long long c_val = suff_cost[idx];
            if (best_suffix_cost.find(s_val) == best_suffix_cost.end() || c_val < best_suffix_cost[s_val]) {
                best_suffix_cost[s_val] = c_val;
            }
        }
        
        long long need = required_sum — pref_sum[p];
        if (best_suffix_cost.find(need) != best_suffix_cost.end()) {
            long long candidate = pref_cost[p] + best_suffix_cost[need];
            min_total_cost = min(min_total_cost, candidate);
        }
    }
    
    if (min_total_cost >= INF) {
        cout << -1 << "\n";
    } else {
        cout << min_total_cost << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int test_cases;
    cin >> test_cases;
    while (test_cases--) {
        solve();
    }
    return 0;
}

Problem Summary & Key Intuition:

• Summary: For each energy value A, find the smallest k >= 0 such that adding k^2 yields a perfect square. If impossible, set A = 1 with k = 0. Output max k used and modified array.
• Key Intuition: We need M^2 — k^2 = A <=> (M — k)(M + k) = A. Setting u = M — k and v = M + k, u * v = A with u and v having the same parity. Minimizing k = (v — u) / 2 requires choosing divisor u <= sqrt(A) as close to sqrt(A) as possible. If A % 4 == 2, no matching-parity factors exist, forcing A -> 1.

Detailed Explanation & Complexity:

• Detailed Explanation: Scan u downwards from floor(sqrt(A)) to 1. The first divisor found where A / u has matching parity gives the minimum gap v — u and thus minimal k.
• Time Complexity: O(N sqrt(max A)) time.
• Space Complexity: O(N) space.

C++ Implementation:

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

using namespace std;

pair<long long, long long> solve_single(long long a) {
    long long sq = round(sqrt((double)a));
    if (sq * sq == a) {
        return {0, a};
    }
    if (a % 2 != 0 && (a % 4 == 2)) {
        return {0, 1};
    }
    if (a % 4 == 2) {
        return {0, 1};
    }
    
    long long best_u = -1;
    for (long long u = sq; u >= 1; u--) {
        if (a % u == 0) {
            long long v = a / u;
            if ((v — u) % 2 == 0) {
                best_u = u;
                break;
            }
        }
    }
    
    if (best_u == -1) {
        return {0, 1};
    }
    
    long long v = a / best_u;
    long long k = (v — best_u) / 2;
    long long m = (v + best_u) / 2;
    return {k, m * m};
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    if (!(cin >> n)) return 0;
    
    vector<long long> final_energies(n);
    long long max_k = 0;
    
    for (int i = 0; i < n; i++) {
        long long a;
        cin >> a;
        auto [k, energy] = solve_single(a);
        max_k = max(max_k, k);
        final_energies[i] = energy;
    }
    
    cout << max_k << "\n";
    for (int i = 0; i < n; i++) {
        cout << final_energies[i] << (i == n — 1 ? "" : " ");
    }
    cout << "\n";
    
    return 0;
}

Problem Summary & Key Intuition:

• Summary: Split binary string S of length N into K non-empty contiguous parts (each length <= 31) to maximize sum(i * value(P_i)).
• Key Intuition: Because each part length is bounded by 31 bits, we can use interval dynamic programming with memoization dp(pos, parts_left) where each state transitions into at most 31 choices.

Detailed Explanation & Complexity:

• Detailed Explanation: Base case dp(N, 0) = 0. For state (pos, parts_left), iterate ending index end from pos up to min(N-1, pos + 30). Compute binary integer value iteratively and transition to dp(end + 1, parts_left — 1).
• Time Complexity: O(31 * N * K) total operations.
• Space Complexity: O(N * K) space for memo table.

C++ Implementation:

 #include <iostream>
 #include <vector>
 #include <string>
 #include <algorithm>
 #include <climits>

using namespace std;

const long long INF = 1e18;

long long solve_dp(const string &s, int pos, int parts_left, int total_k, vector<vector<long long>> &dp) {
    int n = s.length();
    if (pos == n) {
        return parts_left == 0 ? 0 : -INF;
    }
    if (n — pos < parts_left || parts_left <= 0) {
        return -INF;
    }
    if (dp[pos][parts_left] != -1) {
        return dp[pos][parts_left];
    }
    
    int part_number = total_k — parts_left + 1;
    long long best = -INF;
    long long val = 0;
    
    for (int end = pos; end < min(n, pos + 31); end++) {
        val = (val << 1) + (s[end] — '0');
        long long rest = solve_dp(s, end + 1, parts_left — 1, total_k, dp);
        if (rest != -INF) {
            best = max(best, part_number * val + rest);
        }
    }
    
    return dp[pos][parts_left] = best;
}

void solve() {
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    
    vector<vector<long long>> dp(n + 1, vector<long long>(k + 1, -1));
    long long ans = solve_dp(s, 0, k, k, dp);
    cout << ans << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int test_cases;
    cin >> test_cases;
    while (test_cases--) {
        solve();
    }
    return 0;
}

Problem Summary & Key Intuition:

• Summary: Grid pathfinding from top-left to bottom-right. Pick one light frequency from sorted base array B shifted by cell energy G_{r,c}. Chosen frequencies must strictly increase along path. Minimize total cost sum(10^6 — F) or print DOOMED.
• Key Intuition: Maintain running prefix minimum arrays for candidate light frequencies. Monotonic two-pointer updates allow finding the best strictly smaller frequency predecessor from the top or left cell in amortized O(1) time.

Detailed Explanation & Complexity:

• Detailed Explanation: Sort base frequencies B ascending. Iterating through cells (r, c), maintain idx_top and idx_left pointers advancing monotonically to query pref[c][idx_top — 1] and pref[c — 1][idx_left — 1]. Use rolling 1D DP arrays over grid rows to optimize memory.
• Time Complexity: O(N * M * K) total time.
• Space Complexity: O(M * K) memory using rolling arrays.

C++ Implementation:

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

using namespace std;

const long long INF = 1e18;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, m, k;
    if (!(cin >> n >> m >> k)) return 0;
    
    vector<long long> b(k);
    for (int i = 0; i < k; i++) {
        cin >> b[i];
    }
    sort(b.begin(), b.end());
    
    vector<long long> prev_grid(m), curr_grid(m);
    vector<vector<long long>> prev_pref(m, vector<long long>(k, INF));
    vector<vector<long long>> curr_pref(m, vector<long long>(k, INF));
    
    for (int r = 0; r < n; r++) {
        for (int c = 0; c < m; c++) {
            cin >> curr_grid[c];
            long long g_curr = curr_grid[c];
            long long g_top = (r > 0) ? prev_grid[c] : 0;
            long long g_left = (c > 0) ? curr_grid[c — 1] : 0;
            
            long long running_min = INF;
            int idx_top = 0, idx_left = 0;
            
            for (int i = 0; i < k; i++) {
                long long f = g_curr + b[i];
                
                if (r > 0) {
                    while (idx_top < k && g_top + b[idx_top] < f) {
                        idx_top++;
                    }
                }
                if (c > 0) {
                    while (idx_left < k && g_left + b[idx_left] < f) {
                        idx_left++;
                    }
                }
                
                long long best_prev = INF;
                if (r == 0 && c == 0) {
                    best_prev = 0;
                } else {
                    if (r > 0 && idx_top > 0) {
                        best_prev = min(best_prev, prev_pref[c][idx_top — 1]);
                    }
                    if (c > 0 && idx_left > 0) {
                        best_prev = min(best_prev, curr_pref[c — 1][idx_left — 1]);
                    }
                }
                
                if (best_prev != INF) {
                    running_min = min(running_min, best_prev + (1000000LL — f));
                }
                curr_pref[c][i] = running_min;
            }
        }
        
        for (int c = 0; c < m; c++) {
            prev_grid[c] = curr_grid[c];
            for (int i = 0; i < k; i++) {
                prev_pref[c][i] = curr_pref[c][i];
            }
        }
    }
    
    long long ans = prev_pref[m — 1][k — 1];
    if (ans >= INF) {
        cout << "DOOMED\n";
    } else {
        cout << ans << "\n";
    }
    
    return 0;
}

Problem Summary & Key Intuition:

• Summary: Calculate range sums of distinct palindromic subsequence counts f_x(i) over Pascal matrix rows i in [L, R] truncated at column x modulo 998244353.
• Key Intuition: The structural behavior of row i splits i into three main mathematical regions relative to limit x:
  1. Region 1 (i <= x): Full symmetric row, geometric sums of powers of 2 based on even/odd parity.
  2. Region 2 (x < i <= 2x+1): Partially folded row, capped symmetric pair counts.
  3. Region 3 (i >= 2x+2): Strictly increasing window, constant x + 1 distinct elements per row.

Detailed Explanation & Complexity:

• Detailed Explanation: Evaluate contributions from each region using logarithmic power function power_two and range exponent sum formula sum_pow.
• Time Complexity: O(Q log MOD) time.
• Space Complexity: O(1) space.

C++ Implementation:

 #include <iostream>
 #include <algorithm>

using namespace std;

const long long MOD = 998244353;

long long power_two(long long exp) {
    long long res = 1;
    long long base = 2;
    while (exp > 0) {
        if (exp % 2 == 1) res = (res * base) % MOD;
        base = (base * base) % MOD;
        exp /= 2;
    }
    return res;
}

long long sum_pow(long long low, long long high) {
    if (low > high) return 0;
    long long val_high = power_two(high + 1);
    long long val_low = power_two(low);
    return (val_high — val_low + MOD) % MOD;
}

void solve_query() {
    long long l, r, x;
    if (!(cin >> l >> r >> x)) return;
    
    long long total = 0;
    
    long long r1_start = l, r1_end = min(r, x);
    if (r1_start <= r1_end) {
        long long res = 0;
        
        long long even_start = (r1_start % 2 == 0) ? r1_start : r1_start + 1;
        long long even_end = (r1_end % 2 == 0) ? r1_end : r1_end — 1;
        if (even_start <= even_end) {
            long long count = (even_end — even_start) / 2 + 1;
            long long sp = sum_pow(even_start / 2, even_end / 2);
            res = (res + 3 * sp % MOD * (count % MOD)) % MOD;
        }
        
        long long odd_start = (r1_start % 2 != 0) ? r1_start : r1_start + 1;
        long long odd_end = (r1_end % 2 != 0) ? r1_end : r1_end — 1;
        if (odd_start <= odd_end) {
            long long count = (odd_end — odd_start) / 2 + 1;
            long long sp = sum_pow((odd_start + 1) / 2, (odd_end + 1) / 2);
            res = (res + 2 * sp % MOD * (count % MOD)) % MOD;
        }
        
        total = (total + res) % MOD;
    }
    
    long long r2_start = max(l, x + 1), r2_end = min(r, 2 * x + 1);
    if (r2_start <= r2_end) {
        long long res = 0;
        
        long long even_start = (r2_start % 2 == 0) ? r2_start : r2_start + 1;
        long long even_end = (r2_end % 2 == 0) ? r2_end : r2_end — 1;
        if (even_start <= even_end) {
            long long sp = sum_pow(x — even_end / 2, x — even_start / 2);
            res = (res + 3 * sp) % MOD;
        }
        
        long long odd_start = (r2_start % 2 != 0) ? r2_start : r2_start + 1;
        long long odd_end = (r2_end % 2 != 0) ? r2_end : r2_end — 1;
        if (odd_start <= odd_end) {
            long long sp = sum_pow(x — (odd_end + 1) / 2, x — (odd_start + 1) / 2);
            res = (res + 2 * sp) % MOD;
        }
        
        total = (total + res) % MOD;
    }
    
    long long r3_start = max(l, 2 * x + 2), r3_end = r;
    if (r3_start <= r3_end) {
        long long count = r3_end — r3_start + 1;
        total = (total + (x + 1) % MOD * (count % MOD)) % MOD;
    }
    
    cout << total << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int q;
    if (cin >> q) {
        while (q--) {
            solve_query();
        }
    }
    return 0;
}