# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
1250215 | ezzzay | Triple Peaks (IOI25_triples) | C++20 | 0 ms | 0 KiB |
#include "triples.h"
#include <vector>
using namespace std;
// Part I: O(N) scan by anchoring on each index
long long count_triples(const vector<int>& H) {
int N = H.size();
long long ans = 0;
// Case A: H[i] = d(i,j), H[j] = d(j,k), so j=i+H[i], k=j+H[j], and H[k]==H[i]+H[j]
for(int i=0; i<N; i++){
int j = i + H[i];
if(j < N){
int k = j + H[j];
if(k < N && H[k] == H[i] + H[j])
ans++;
}
}
// Case B: H[i] = d(i,j), H[k] = d(i,k), so k=i+H[i], j=k-H[k], and H[j]==H[i]+H[k]
for(int i=0; i<N; i++){
int k = i + H[i];
if(k < N){
int j = k - H[k];
if(j > i && j < N && H[j] == H[i] + H[k])
ans++;
}
}
// Case C: H[j] = d(j,k), H[j] = d(i,j), so i=j-H[j], k=j+H[j], and H[i]+H[k]==H[j]
for(int j=0; j<N; j++){
int i = j - H[j];
int k = j + H[j];
if(i >= 0 && k < N && H[i] + H[k] == H[j])
ans++;
}
return ans;
}
// Part II: fill with 1s → every consecutive triple (i,i+1,i+2) is mythical.
// You get (N-2) triples out of N peaks.
vector<int> construct_range(int M, int K) {
int N;
if(M >= K + 2) {
// can hit exactly K triples → full credit
N = K + 2;
} else if(M >= 3) {
// use all M peaks → M-2 triples
N = M;
} else {
// too small for any triple, just output M ones
N = M;
}
return vector<int>(N, 1);
}