#include "triples.h"
#define ll long long
using namespace std;
long long count_triples(std::vector<int> H) {
//firstly, consider index 1 is part of a mythical triple (1 is a label not an index)
//there are two possibilities, either that index 1 and 2 are separated by the height of 1 (or index 1 and 3)
//or that indices 2 and 3 are separated by the height of 1
//if it's the former case, it's simple because we can just check index 2 positions +- H_1 of position of index 1
//and then check index 3 positions +- H_2 of position of index 1 and then +- H_2 of position of index 2
//so only 2 index 2 and 4 index 3 positions need to be considered
//in the case where indices 2 and 3 are separated by the height of 1
//then if indices 1 and 2 are separated by the height of 2 we're essentially back to the first case
//the only really hard case is if indices 1 and 2 are separated by the height of 3
//and indices 1 and 3 are separated by the height of 2
//etc
//ok note that one of the pairwise distances is the sum of the other two
//and thus one of the heights is the sum of the other two
//in fact the middle height is the sum of the 2 outer ones
//which means this case is impossible in subtask 4
//so we can easily get 35 points with this solution so far
//so we want two indices such that the distance from the first to index 1 on the left is equal to the height of the second
//and the distance from index 1 on the right to the second is the height of the first
//call these indices a,b and the middle index M
//b = a+H[M]
//so we definitely need:
//M-a = H[b]
//b-M = H[a]
//so summing, b-a = H[b]+H[a]
//maybe we first find pairs of indices such that b-a = H[b]+H[a] and then check that their midpoint works
//in fact this means that H[a]+a = b-H[b]
//so we can look at the values of H[a]+a over all a, look at the values of b-H[b] over all b, and then see for overlaps
return 0ll;
}
std::vector<int> construct_range(int M, int K) {
int block_size = 5;
int series = 50;
vector<int> result;
for (int i=series-1; i>=0; i--) {
for (int j=1; j<=block_size; j++) {
result.push_back(i*block_size+j);
}
}
for (int j=block_size-1; j>=1; j--) {
result.push_back(j);
}
return result;
}
