D - Max Straight

This problem can be solved efficient using a hash map. Initially, I overcomplicated it by thinking about sorting and finding the longest increasing subsequence. However, a simple approach using a map workss perfectly.

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    unordered_map<int, int> freq;
    int result = 0;
    
    for (int i = 0; i < n; ++i) {
        int value;
        cin >> value;
        freq[value] = max(freq[value], freq[value - 1] + 1);
        result = max(result, freq[value]);
    }
    
    cout << result << endl;
    return 0;
}

E - Multiple-Free Sequences

This is a mathematical problem involving cycle detection in modular arithmetic. A depth-first search (DFS) is used to traverse sequences and determine which pairs are valid.

int answer = 0;
bool visited[100005][100005];
bool invalid[100005][100005];

void dfs(int x, int y) {
    if (x == 0 || y == 0) {
        invalid[x][y] = true;
        return;
    }
    
    if (visited[x][y]) return;
    visited[x][y] = true;
    
    int next_x = y % m;
    int next_y = (a * y + b * x) % m;
    dfs(next_x, next_y);
    
    invalid[x][y] = invalid[next_x][next_y];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    cin >> m >> a >> b;
    
    for (int x = 1; x < m; ++x) {
        for (int y = 1; y < m; ++y) {
            if (!visited[x][y]) dfs(x, y);
            answer += !invalid[x][y];
        }
    }
    
    cout << answer;
    return 0;
}

F - Reachable Set 2

This problem involves determining reachabliity in a directed graph using shortest path algorithms and then calculating how many nodes can be removed while maintaining reachability. The key steps involve running a modified Dijkstra’s algorithm and tracking prefix maximums.

#define N 100005
#define pii pair<int, int>

int n, m;
vector<int> adj[N];
priority_queue<pii, vector<pii>, greater<pii> > pq;
bool visited[N];
int distance[N];
int prefix_max[N];
map<pii, int> edge_map;
bool marked[N];

void dijkstra() {
    memset(distance, 0x3f, sizeof distance);
    distance[1] = 1;
    pq.push({1, 1});
    
    while (!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        
        if (visited[u]) continue;
        visited[u] = true;
        
        for (auto v : adj[u]) {
            if (distance[v] > max(distance[u], v)) {
                distance[v] = max(distance[u], v);
                pq.push({distance[v], v});
            }
        }
    }
}

void solve() {
    cin >> n >> m;
    int u, v;
    
    for (int i = 1; i <= m; ++i) {
        cin >> u >> v;
        if (u == v) continue;
        if (edge_map[{u, v}]) continue;
        adj[u].push_back(v);
        edge_map[{u, v}] = 1;
    }
    
    dijkstra();
    
    for (int i = 1; i <= n; ++i) {
        prefix_max[i] = max(prefix_max[i - 1], distance[i]);
    }
    
    int count = 0;
    
    for (int i = 1; i <= n; ++i) {
        bool flag = false;
        
        for (auto node : adj[i]) {
            if (!marked[node]) {
                marked[node] = true;
                count++;
            }
        }
        
        if (marked[i]) count--;
        marked[i] = true;
        
        if (prefix_max[i] > i) cout << -1 << endl;
        else cout << count << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    while (T--) {
        solve();
    }
    return 0;
}