The problem involved a simulation where characters 'a', 'b', and 'c' each appeared exactly once. The initial misunderstanding of the problem statement led to multiple incorrect attempts. The key was recognizing that each character appeared only once, not at least once.

Problem E: Expected Value Calculation

To solve the expected value problem, we used the formula ans = Σ(a_i × p_i), where p_i represnets the expected number of times we need to roll the dice to get value i.

The relationship p_i = (1/n) × Σ(p_j for j from 0 to i-1) allows for linear computation of the solution.

#include <bits>
using namespace std;

const int MOD = 1000000007;
const int MAXN = 100005;

int n, values[MAXN];

// Modular exponentiation
long long mod_pow(long long base, long long exp) {
    long long result = 1;
    while (exp > 0) {
        if (exp & 1) 
            result = (result * base) % MOD;
        base = (base * base) % MOD;
        exp >>= 1;
    }
    return result;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> values[i];
    }
    
    long long inv_n = mod_pow(n, MOD - 2);
    long long result = 0;
    long long current_prob = inv_n;
    
    for (int i = 1; i <= n; ++i) {
        result = (result + current_prob * values[i]) % MOD;
        current_prob = (current_prob + current_prob * inv_n) % MOD;
    }
    
    cout << result << "\n";
    return 0;
}
</bits>

Problem F: Optimized Search with X-Y Axis Separation

Given the constraint n ≤ 80, a brute-force aprpoach wasn't feasible. The solution involved separating the x and y axes calculations, reducing complexity to 2 × 2^40. Further optimization used bidirectional search to handle the initial and final states efficiently.

The conversion from axis positions to directional moves (L/R) required tracking previous direction information.

#include <bits>
using namespace std;

const int MAXN = 85;

int n, start_x, start_y;
vector<int> x_moves, y_moves;
map<int string=""> x_paths, y_paths;
map<int int=""> visited;

void dfs_x(int idx, int pos, string path) {
    if (idx == x_moves.size() / 2) {
        x_paths[pos] = path;
        return;
    }
    dfs_x(idx + 1, pos + x_moves[idx], path + "1");
    dfs_x(idx + 1, pos - x_moves[idx], path + "0");
}

bool found = false;
void dfs_x_rev(int idx, int pos, string path) {
    if (idx == x_moves.size()) {
        if (!x_paths.count(start_x - pos)) return;
        x_paths[start_x] = x_paths[start_x - pos] + path;
        found = true;
        return;
    }
    dfs_x_rev(idx + 1, pos + x_moves[idx], path + "1");
    if (found) return;
    dfs_x_rev(idx + 1, pos - x_moves[idx], path + "0");
    if (found) return;
}

void dfs_y(int idx, int pos, string path) {
    if (idx == y_moves.size() / 2) {
        y_paths[pos] = path;
        return;
    }
    dfs_y(idx + 1, pos + y_moves[idx], path + "1");
    dfs_y(idx + 1, pos - y_moves[idx], path + "0");
}

bool found_y = false;
void dfs_y_rev(int idx, int pos, string path) {
    if (idx == y_moves.size()) {
        if (!y_paths.count(start_y - pos)) return;
        y_paths[start_y] = y_paths[start_y - pos] + path;
        found_y = true;
        return;
    }
    dfs_y_rev(idx + 1, pos + y_moves[idx], path + "1");
    if (found_y) return;
    dfs_y_rev(idx + 1, pos - y_moves[idx], path + "0");
    if (found_y) return;
}

void construct_solution() {
    cout << "Yes\n";
    vector<int> combined;
    int i = 0, j = 0;
    
    while (i < y_paths[start_y].size() || j < x_paths[start_x].size()) {
        if (i < y_paths[start_y].size()) {
            combined.push_back(y_paths[start_y][i] - '0');
            ++i;
        }
        if (j < x_paths[start_x].size()) {
            combined.push_back(x_paths[start_x][j] - '0');
            ++j;
        }
    }
    
    int last_dir = 0;
    for (int move : combined) {
        if (move) {
            if (last_dir == 0) cout << "L", last_dir = 1;
            else if (last_dir == 1) cout << "R", last_dir = 0;
            else if (last_dir == 2) cout << "R", last_dir = 1;
            else if (last_dir == 3) cout << "L", last_dir = 0;
        } else {
            if (last_dir == 0) cout << "R", last_dir = 3;
            else if (last_dir == 1) cout << "L", last_dir = 2;
            else if (last_dir == 2) cout << "L", last_dir = 3;
            else if (last_dir == 3) cout << "R", last_dir = 2;
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n >> start_x >> start_y;
    for (int i = 1; i <= n; ++i) {
        int val;
        cin >> val;
        if (i % 2 == 1) y_moves.push_back(val);
        else x_moves.push_back(val);
    }
    
    dfs_x(0, 0, "");
    dfs_x_rev(x_moves.size() / 2, 0, "");
    
    found_y = false;
    dfs_y(0, 0, "");
    dfs_y_rev(y_moves.size() / 2, 0, "");
    
    if (x_paths.count(start_x) && y_paths.count(start_y)) {
        construct_solution();
    } else {
        cout << "No\n";
    }
    
    return 0;
}
</int></int></int></int></bits>

Problem G: Network Flow with Minimum Cut

The problem required constructing a network flow graph to find the minimum cut. With L_{i,j} ≤ 5, the number of nodes was limited. The solution involved adding all values and calculating the minimum cut, with special attention to the constraints between nodes of different levels.

Nodes that didn't satisfy conditions were connected with infinite capacity edges to higher-level nodes. Similarly, conditions that granted value despite not being met were connected with infinite capacity edges.

#include <bits>
using namespace std;

const int MAXN = 1005;
const int INF = 1e9;

int n, m;
int costs[MAXN], achievements[MAXN];

struct MaxFlow {
    struct Edge {
        int to, rev, cap;
    };
    
    vector<edge> graph[MAXN];
    int level[MAXN], iter[MAXN];
    
    void add_edge(int from, int to, int cap) {
        graph[from].push_back({to, (int)graph[to].size(), cap});
        graph[to].push_back({from, (int)graph[from].size() - 1, 0});
    }
    
    void bfs(int s) {
        memset(level, -1, sizeof(level));
        queue<int> q;
        level[s] = 0;
        q.push(s);
        
        while (!q.empty()) {
            int v = q.front();
            q.pop();
            
            for (auto &e : graph[v]) {
                if (e.cap > 0 && level[e.to] < 0) {
                    level[e.to] = level[v] + 1;
                    q.push(e.to);
                }
            }
        }
    }
    
    int dfs(int v, int t, int f) {
        if (v == t) return f;
        
        for (int &i = iter[v]; i < graph[v].size(); ++i) {
            Edge &e = graph[v][i];
            
            if (e.cap > 0 && level[v] < level[e.to]) {
                int d = dfs(e.to, t, min(f, e.cap));
                if (d > 0) {
                    e.cap -= d;
                    graph[e.to][e.rev].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }
    
    int max_flow(int s, int t) {
        int flow = 0;
        while (true) {
            bfs(s);
            if (level[t] < 0) return flow;
            
            memset(iter, 0, sizeof(iter));
            int f;
            while ((f = dfs(s, t, INF)) > 0) {
                flow += f;
            }
        }
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n >> m;
    MaxFlow flow;
    int source = 5 * n + m, sink = 5 * n + m + 1;
    int total_value = 0;
    
    for (int i = 0; i < n; ++i) {
        cin >> costs[i];
        for (int j = 0; j < 4; ++j) {
            flow.add_edge(5 * i + j, 5 * i + j + 1, costs[i] * j);
            flow.add_edge(5 * i + j + 1, 5 * i + j, INF);
        }
        flow.add_edge(5 * i + 4, sink, costs[i] * 4);
    }
    
    for (int i = 1; i <= m; ++i) {
        cin >> achievements[i];
        total_value += achievements[i];
    }
    
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            int requirement;
            cin >> requirement;
            --requirement;
            flow.add_edge(5 * n + i, 5 * j + requirement, INF);
        }
        flow.add_edge(source, 5 * n + i, achievements[i + 1]);
    }
    
    total_value -= flow.max_flow(source, sink);
    cout << total_value << "\n";
    
    return 0;
}
</int></edge></bits>