Overview
This contest proved challenging despite seemingly moderate difficulty. The overall rating leans toward green to purple, but the execution was frustrating. T1 cost me significant points due to rushing through it—225 dropped to 175 points. Strategic lesson: even when T1 appears simple, allocating proper time (up to 1.5 hours is reasonable) remains crucial.
Problem 1: Arithmetic Operations (add)
This problem originated from Codeforces (440C). The challenge involves finding the minimum count of ones needed to construct a given number n, where numbers are formed by concatenating repeated digit 1.
A key insight: always use the longest possible number at each step. This greedy approach works because longer sequences of ones become increasingly efficient. A depth-first search explores the solution space.
// Arithmetic problem - finding minimum ones
#include <bits/stdc++.h>
using namespace std;
long long target;
long long powerOfTen[20], repunit[20];
void precompute()
{
repunit[1] = powerOfTen[1] = 1;
for (int i = 2; i <= 16; ++i)
{
powerOfTen[i] = powerOfTen[i - 1] * 10;
repunit[i] = repunit[i - 1] * 10 + 1;
}
}
int minOperations = INT_MAX;
void explore(long long current, int digitPos, int ops)
{
if (ops > minOperations) return;
if (current == 0) { minOperations = min(minOperations, ops); return; }
if (current < 0) current = -current;
if (current / powerOfTen[digitPos] == 0)
{
explore(current, digitPos - 1, ops);
return;
}
// Strategy 1: subtract repunit
long long val1 = current;
int addOps1 = 0;
while (val1 >= powerOfTen[digitPos])
{
val1 -= repunit[digitPos];
addOps1 += digitPos;
}
explore(val1, digitPos - 1, ops + addOps1);
// Strategy 2: go to next repunit and subtract
long long val2 = repunit[digitPos + 1] - current;
int addOps2 = digitPos + 1;
while (val2 >= powerOfTen[digitPos])
{
val2 -= repunit[digitPos];
addOps2 += digitPos;
}
explore(val2, digitPos - 1, ops + addOps2);
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
precompute();
cin >> target;
explore(target, 15, 0);
cout << minOperations << '\n';
return 0;
}
Problem 2: Escape Route (car)
This green-level problem examines vehicle departure sequences. The core insight: whether a vehicle can depart depends not only on its immediate predecessor but propagates through the entire preceding sequence.
When a vehicle departs, it affects all subsequent vehicles. Only when reaching a certain critical position does a vehicle cease influencing those behind it. A priority queue efficiently manages these dependencies.
// Vehicle departure simulation
#include <bits/stdc++.h>
using namespace std;
struct Vehicle {
long long start, duration, speed;
};
const int MAXN = 300005;
int totalVehicles, targetTime, operations, boost;
Vehicle vehicles[MAXN];
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> totalVehicles >> targetTime >> operations >> boost;
for (int i = 1; i <= totalVehicles; ++i)
cin >> vehicles[i].start >> vehicles[i].duration >> vehicles[i].speed;
// Sort by start position
sort(vehicles + 1, vehicles + totalVehicles + 1,
[](const Vehicle& a, const Vehicle& b) { return a.start < b.start; });
// Calculate accumulated time
vector<long long> prefixSum(totalVehicles + 1);
for (int i = 1; i <= totalVehicles; ++i)
prefixSum[i] = prefixSum[i - 1] + vehicles[i].duration;
using pdi = pair<double, int>;
priority_queue<pdi> pq;
for (int i = 1; i <= totalVehicles; ++i)
{
double required = 1.0 * (prefixSum[i - 1] + targetTime - vehicles[i].start) / vehicles[i].speed;
pq.push({required, i});
}
for (int i = 0; i < operations; ++i)
{
auto current = pq.top(); pq.pop();
int idx = current.second;
vehicles[idx].speed += boost;
double newRequired = 1.0 * (prefixSum[idx - 1] + targetTime - vehicles[idx].start) / vehicles[idx].speed;
pq.push({newRequired, idx});
}
cout << fixed << setprecision(3) << pq.top().first << '\n';
return 0;
}
Problem 3: Yugoslavian Problem (yugo)
This blue-level graph theory problem involves constructing a spanning tree and determining edge directions. The solution builds three graphs: the original undirected graph, the spanning tree, and the directed answer graph.
The approach processes the original graph to extract the tree and partially construct the answer, then traverses the tree to finalize edge directions. Using namespaces significantly reduces variable collision bugs.
// Yugoslavian problem - graph construction and edge direction
#include <bits/stdc++.h>
using namespace std;
struct EdgeList {
struct Node { int dest, next, edgeId; };
vector<Node> edges;
vector<int> head, degree;
int edgeCount;
EdgeList(int n = 0) { initialize(n); }
void initialize(int n) {
edges.clear(); edges.reserve(2 * n);
head.assign(n + 1, 0);
degree.assign(n + 1, 0);
edgeCount = 0;
}
void addEdge(int from, int to, int id) {
degree[to]++;
edges.push_back({to, head[from], id});
head[from] = (int)edges.size() - 1;
}
};
const int MAXN = 1000005;
const char UNSTABLE_MSG[] = "Yugoslavia is unstable!";
int testCases, vertexCount, edgeCount;
pair<int, int> edgeEndpoints[MAXN];
bool validSolution = true;
EdgeList original, spanning, result;
vector<bool> inSpanning(2 * MAXN), visitedVertex(MAXN), processed(2 * MAXN);
inline int reverseEdge(int id) { return (id & 1) ? id + 1 : id - 1; }
vector<int> allocatedCount;
namespace TreeBuilder {
int traversalCount;
void dfs(int vertex)
{
visitedVertex[vertex] = true;
for (int i = original.head[vertex]; i; i = original.edges[i].next)
{
int neighbor = original.edges[i].dest;
int eid = original.edges[i].edgeId;
traversalCount++;
if (visitedVertex[neighbor])
{
if (!inSpanning[i] && !processed[i])
{
processed[i] = processed[reverseEdge(i)] = true;
result.addEdge(vertex, neighbor, eid);
}
}
else
{
inSpanning[i] = inSpanning[reverseEdge(i)] = true;
spanning.addEdge(vertex, neighbor, eid);
spanning.addEdge(neighbor, vertex, eid);
dfs(neighbor);
}
}
}
void processAll()
{
for (int i = 1; i <= vertexCount; ++i)
{
if (!visitedVertex[i])
{
traversalCount = 0;
dfs(i);
if ((traversalCount >> 1) & 1)
{
validSolution = false;
return;
}
}
}
}
}
namespace DirectionSetter {
vector<bool> marked;
void dfs(int vertex)
{
marked[vertex] = true;
for (int i = spanning.head[vertex]; i; i = spanning.edges[i].next)
{
int neighbor = spanning.edges[i].dest;
int eid = spanning.edges[i].edgeId;
if (marked[neighbor]) continue;
dfs(neighbor);
if (result.degree[neighbor] & 1)
result.addEdge(vertex, neighbor, eid);
else
result.addEdge(neighbor, vertex, eid);
}
}
void processAll()
{
marked.assign(vertexCount + 1, false);
for (int i = 1; i <= vertexCount; ++i)
if (!marked[i]) dfs(i);
}
}
vector<int> finalOutput;
void constructOutput()
{
for (int i = 1; i <= result.edgeCount; ++i)
{
int dest = result.edges[i].dest;
int eid = result.edges[i].edgeId;
if (dest == edgeEndpoints[eid].first)
{
if (++allocatedCount[dest] <= result.degree[dest] / 2)
finalOutput[eid] = 1;
else
finalOutput[eid] = 3;
}
else
{
if (++allocatedCount[dest] <= result.degree[dest] / 2)
finalOutput[eid] = 2;
else
finalOutput[eid] = 4;
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> testCases >> vertexCount >> edgeCount;
original.initialize(vertexCount);
spanning.initialize(vertexCount);
result.initialize(vertexCount);
for (int i = 1; i <= edgeCount; ++i)
{
cin >> edgeEndpoints[i].first >> edgeEndpoints[i].second;
original.addEdge(edgeEndpoints[i].first, edgeEndpoints[i].second, i);
original.addEdge(edgeEndpoints[i].second, edgeEndpoints[i].first, i);
}
TreeBuilder::processAll();
if (validSolution)
{
DirectionSetter::processAll();
allocatedCount.assign(vertexCount + 1, 0);
finalOutput.assign(edgeCount + 1, 0);
constructOutput();
for (int i = 1; i <= edgeCount; ++i)
cout << finalOutput[i] << ' ';
}
else
{
cout << UNSTABLE_MSG;
}
return 0;
}
Problem 4: Counting Problem (count)
This green-level mathematical problem involves algebraic manipulation. The key transformasions involve substituting x = a_i + 1 and multiplying both sides by (x + y).
Essantial knowledge: the cubic difference factorization formula x³ - y³ = (x - y)(x² + xy + y²). An unordered map efficiently tracks value frequencies.
// Counting problem - mathematical transformation
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500005;
int numElements, modulus;
long long values[MAXN];
unordered_map<int, int> countX, countY;
inline long long cubic(long long v) { return (__int128)v * v * v; }
inline long long square(long long v) { return (__int128)v * v; }
inline int normalize(long long v) {
v %= modulus;
if (v < 0) v += modulus;
return (int)v;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> numElements >> modulus;
long long answer = 0;
for (int i = 1; i <= numElements; ++i)
{
cin >> values[i];
long long base = values[i] + 1;
int leftVal = normalize(cubic(base) - square(base));
int rightVal = normalize(-cubic(values[i]) - square(values[i]));
answer += countY[leftVal];
countY[rightVal]++;
answer += countX[rightVal];
countX[leftVal]++;
}
cout << answer << '\n';
return 0;
}
Key Takeaways
- Never rush T1—a careful approach within 1.5 hours is reasonable even to seemingly simple problems
- When algorithm correctness is uncertain, always implement brute force for comparison testing
- Maintain proper contest strategy rather than aggressive time allocation
- Essential mathematical formulas: cubic difference expansions and repunit number properties