Problem A: Large-Scale Simulation

A pure simulation problem centered on game theory mechanics. The implementation involves directly modeling the described rules and state transitions.

Problem B: Maximum Total Range for k-Partition

Define the weight of a subarray as its range (maximum element minus minimum element). For each k from 1 to n, compute the maximum possible sum of weights when partitioning the array into exactly k contiguous segments.

Constraints: n ≤ 5000, element values ≤ 10⁴

Dynamic Programming Approach:

Let dp[pos][seg][mask] represent processing the first pos elements using seg segments, where mask (0-3) tracks whether the current segment has captured its minimum and maximum values. The solution uses a rolling array to optimize space to O(n²).

const int MAXN = 5010;
int arr[MAXN], dp[2][MAXN][4];

int max_of_three(int p, int q, int r) { return std::max(std::max(p, q), r); }
int max_of_four(int a, int b, int c, int d) { return std::max(std::max(a, b), std::max(c, d)); }

int main() {
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
    
    memset(dp, -0x3f, sizeof(dp));
    dp[0][0][3] = 0;
    
    int cur = 0, prev = 1;
    for (int i = 1; i <= n; ++i) {
        std::swap(cur, prev);
        for (int j = 1; j <= n; ++j) {
            dp[cur][j][0] = std::max(dp[prev][j-1][3], dp[prev][j][0]);
            dp[cur][j][1] = max_of_three(dp[prev][j-1][3] + arr[i], dp[prev][j][0] + arr[i], dp[prev][j][1]);
            dp[cur][j][2] = max_of_three(dp[prev][j-1][3] - arr[i], dp[prev][j][2], dp[prev][j][0] - arr[i]);
            dp[cur][j][3] = max_of_four(dp[prev][j-1][3], dp[prev][j][3], dp[prev][j][1] - arr[i], dp[prev][j][2] + arr[i]);
        }
    }
    
    for (int k = 1; k <= n; ++k)
        printf("%d\n", dp[cur][k][3]);
    return 0;
}

Problem C: Absolute Difference Pairing

Indices i and j can be matched if |i - a_i| = |j - a_j|. For an even-sized array, determine if a perfect matching exists and output one valid configuraton.

Constraints: T ≤ 10, total n ≤ 10⁶, |a_i| ≤ 10⁹

Graph-Theoretic Construction:

The matching condition implies either i + a_i = j + a_j or i - a_i = j - a_j. Construct a bipartite graph where left nodes represent (i - a_i) and right nodes represent (i + a_i). Each edge must have exactly one endpoint toggled (XOR with 1) to satisfy parity constraints.

For each connected component, extract a spanning tree and perform a post-order traversal to assign values greedily, ensuring all nodes ressolve to zero. This yields a valid perfect matching when one exists.

Problem D: Dynamic Sets with XOR Validation

Maintain a collection of disjoint sets with initial values. Support three operations:

1. Merge sets containing elements x and y
2. Add 1 to every element in x's set
3. Add k to a specific element a_x

After each operation, verify if the XOR sum of all elements in x's set equals zero.

Constraints: n, q ≤ 3×10⁵, values bounded by 10⁹

Binary Trie Solution:

Represent each set as a binary trie where nodes store count and XOR sum of subtrees. This enables efficient bulk operations.

• Merge: Union-Find with small-to-large trie merging
• Global Increment: Recursively swap children and propagate to the left child, mirroring binary addition of 1
• Point Update: Remove the old value (decrement count) and insert the updated value

Complexity: O((n + q) log V) where V is the maximum value bound.

const int MAXN = 300010;
const int MAXNODE = MAXN * 31;
const int MAXBIT = 30;

int child[MAXNODE][2], cnt[MAXNODE], xorSum[MAXNODE], nodeCnt = 0;

int createNode() {
    int id = ++nodeCnt;
    child[id][0] = child[id][1] = cnt[id] = xorSum[id] = 0;
    return id;
}

void update(int id) {
    cnt[id] = xorSum[id] = 0;
    if (child[id][0]) {
        cnt[id] += cnt[child[id][0]];
        xorSum[id] ^= xorSum[child[id][0]] << 1;
    }
    if (child[id][1]) {
        cnt[id] += cnt[child[id][1]];
        xorSum[id] ^= (xorSum[child[id][1]] << 1) | (cnt[child[id][1]] & 1);
    }
}

void insertValue(int &root, int value, int depth, int delta) {
    if (!root) root = createNode();
    if (depth > MAXBIT) { cnt[root] += delta; return; }
    insertValue(child[root][value & 1], value >> 1, depth + 1, delta);
    update(root);
}

int mergeTries(int p, int q) {
    if (!p || !q) return p | q;
    cnt[p] += cnt[q];
    xorSum[p] ^= xorSum[q];
    child[p][0] = mergeTries(child[p][0], child[q][0]);
    child[p][1] = mergeTries(child[p][1], child[q][1]);
    return p;
}

void incrementTrie(int id) {
    std::swap(child[id][0], child[id][1]);
    if (child[id][0]) incrementTrie(child[id][0]);
    update(id);
}

int parent[MAXN], lazyAdd[MAXN], trieRoot[MAXN];
std::set<int> memberSet[MAXN];

int findSet(int x) { return x == parent[x] ? x : parent[x] = findSet(parent[x]); }

int main() {
    int n, q; scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) {
        int val; scanf("%d", &val);
        insertValue(trieRoot[i], val, 0, 1);
        parent[i] = i; lazyAdd[i] = 0;
        memberSet[i].insert(i);
    }
    
    while (q--) {
        int op; scanf("%d", &op);
        if (op == 1) {
            int x = findSet(read()), y = findSet(read());
            if (x == y) continue;
            if (cnt[trieRoot[x]] < cnt[trieRoot[y]]) std::swap(x, y);
            for (int elem : memberSet[y]) {
                memberSet[x].insert(elem);
            }
            trieRoot[x] = mergeTries(trieRoot[x], trieRoot[y]);
            parent[y] = x;
        } else if (op == 2) {
            int x = findSet(read());
            incrementTrie(trieRoot[x]);
            ++lazyAdd[x];
        } else if (op == 3) {
            int x, k; scanf("%d%d", &x, &k);
            int setId = findSet(x);
            insertValue(trieRoot[setId], values[x] + lazyAdd[setId], 0, -1);
            values[x] += k;
            insertValue(trieRoot[setId], values[x] + lazyAdd[setId], 0, 1);
        }
        int x = read();
        printf("%s\n", xorSum[trieRoot[findSet(x)]] ? "Yes" : "No");
    }
    return 0;
}
</int>