| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1251841 | s4dz | Triple Peaks (IOI25_triples) | C++20 | 0 ms | 0 KiB |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
//sub4:
vector<int> construct_range(int M, int K) {
vector<int> ans1 = {
4, 3, 1, 2, 1,
4, 3, 2, 7, 6,
5, 8, 11, 10, 9,
1, 7, 2, 3, 4,
};
if(M <= 20){
while(ans1.size() > M || ans1.back() >= M) ans1.pop_back();
return ans1;
}
vector<int> S = {0};
vector<bool> in_span(2*M, false);
int span_cnt = 0;
vector<pair<int,int>> pts;
while(span_cnt < M-1) {
int best_t = -1, best_gain = -1;
vector<bool> inS(2*M, false);
for(int u:S) inS[u] = true;
for(int t = 0; t <= 2*(M-1); t += 2) if(!inS[t]) {
int gain = 0;
for(int u:S) {
if(u==t) continue;
int s = u + t;
if(s >= 2 && s <= 2*(M-1) && (s%2==0) && !in_span[s]) ++gain;
}
if(gain > best_gain) {
best_gain = gain;
best_t = t;
}
}
if(best_t < 0) break;
for(int u:S) {
if(u == best_t) continue;
int s = u + best_t;
if(s >= 2 && s <= 2*(M-1) && (s%2==0) && !in_span[s]) {
int x = min(u, best_t), y = max(u, best_t);
pts.emplace_back(x, y);
in_span[s] = true;
++span_cnt;
}
}
S.push_back(best_t);
}
vector<int> H(M, 1);
for(auto [x,y] : pts) {
int idx = (x + y) / 2;
int h = (y - x) / 2;
if(idx >= 0 && idx < M) {
H[idx] = h;
}
}
return H;
}
/*long long count_triples(vector<int> H)
{
int n = H.size();
ll ans = 0;
for (int k = 0; k < n; k++)
{
int c = H[k];
int i = k - c;
if (i < 0 || i >= n) continue;
int a = H[i];
if (a > c - a) continue;
int need = c - a;
int j1 = i + a;
int j2 = k - a;
if (i < j1 && j1 < k && H[j1] == need) ans++;
if (j2 != j1 && i < j2 && j2 < k && H[j2] == need) ans++;
}
return ans;
}*/
const int nmax = 2e5 + 5;
unordered_map <int, vector <int>> m;
int n;
ll count_triples(vector <int> H)
{
n = H.size();
ll cnt = 0;
for (int j=0; j<n; j++)
{
int l = j - H[j];
if(l >= 0)
{
int x = H[l];
int y = H[j] - x;
if(y > 0)
{
if(H[l + x] == y) cnt++;
if(y != x && H[l + y] == y) cnt++;
}
}
int r = j + H[j];
if(r < n)
{
int x = H[r];
int y = H[j] - x;
if(y > 0)
{
if(H[j + x] == y) cnt++;
if(y != x && H[j + y] == y) cnt++;
}
}
}
for (int k = 0; k < n; ++k)
{
for (int i : m[k - hk])
{
int j1 = i+H[i];
int j2 = i+H[k];
if (j1<k && H[j1]==k-i) cnt++;
if (j2<k && j2!=j1 && H[j2]==k-i) cnt++;
}
m[k + H[k]].push_back(k);
}
return cnt;
}