Submission #1253034

#	Time	Username	Problem	Language	Result	Execution time	Memory
1253034		walrusramen21	Triple Peaks (IOI25_triples)	C++20	23.29 / 100	2096 ms	2492 KiB

triples

#include "triples.h"
#include <algorithm>
#include <array>
#include <iterator>
#include <cassert>
#include <iostream>

long long count_triples(std::vector<int> H) {
	int N = H.size();
	long long ans = 0;
	// remember we have j-i, k-j, and (j-i) + (k-j)
	for (int i = 0; i < N; i++) {
		for (int j = i+1; j < N; j++) {
			// we have three cases: is H[i], H[j], or H[k] biggest?

			// CASE 1
			// if H[k] is biggest, then either H[i] = j-i and H[j] = k-j
			// or H[i] = k-j and H[j] = j-i
				// if H[i] = j-i and H[j] = k-j, then H[k] = H[i] + H[j]
				// k = (k-j) + (j-i) + i = H[j] + H[i] + i
				// if H[j] = j-i and H[i] = k-j, then still H[k] = H[i] + H[j]
				// and k = H[i] + H[j] + i
			// both sub-cases are the same equation
			int k = H[i] + H[j] + i;
			if (0 <= k && k < N && k != i && k != j && H[k] == H[i] + H[j] && (H[i] == j-i || H[j] == j-i)) {
				// std::cout << i << ' ' << j << ' ' << k << ' ' << "(H[k] is biggest)" << '\n';
				++ans;
			}

			// CASE 2
			// if H[i] is biggest, H[i] = k-i
			// either H[j] = j-i, or H[j] = k-j
				// if H[i] = k-i, H[j] = j-i, H[k] = k-j
				// then H[k] = H[i] - H[j], and k = H[i] - H[j] + j
				// if H[i] = k-i, H[j] = k-j, H[k] = j-i
				// H[k] = H[i] - H[j], k = H[i] + i = H[j] + j
			int k1 = H[i] - H[j] + j;
			int k2 = H[i] + i;
			if (0 <= k1 && k1 < N && k1 != i && k1 != j && H[k1] == H[i] - H[j] && H[j] == j-i) {
				// std::cout << i << ' ' << j << ' ' << k1 << ' ' << "(H[i] is biggest)" << '\n';
				++ans;
			}
			if (k2 != k1 && 0 <= k2 && k2 < N && k2 == H[j] + j && k2 != i && k2 != j && H[k2] == H[i] - H[j] && H[i] == k2-i) {
				// std::cout << i << ' ' << j << ' ' << k2 << ' ' << "(H[i] is biggest)" << '\n';
				++ans;
			}

			// CASE 3
			// if H[j] is biggest, then we got two cases (again!)
				// H[i] = j-i, H[j] = k-i, H[k] = k-j
				// H[k] = H[j] - H[i], k = H[j] + i
				// H[i] = k-j, H[j] = k-i, H[k] = j-i
				// H[k] = H[j] - H[i], k = H[i] + j = H[j] + i
			// note that this means if k exists in the latter case, it should alr satisfy the former case
			k = H[j] + i;
			if (0 <= k && k < N && k != i && k != j && H[k] == H[j] - H[i] && (H[i] == j-i || H[i] == k-j)) {
				// std::cout << i << ' ' << j << ' ' << k << ' ' << "(H[j] is biggest)" << '\n';
				++ans;
			}
		}
	}
	return ans;
}

std::vector<int> construct_range(int M, int K) {
	std::vector<int> ans;
	ans.push_back(1);
	ans.push_back(1);
	for (int i = 2; i < M; i++) {
		ans.push_back(i);
	}
	return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 0 / 6
• Subtask 3: 10 / 10
• Subtask 4: 0 / 11
• Subtask 5: 0 / 16
• Subtask 6: 0 / 19
• Subtask 7: 3 / 5
• Subtask 8: 1.245 / 5
• Subtask 9: 0.4998 / 5
• Subtask 10: 0.2143 / 5
• Subtask 11: 0.25 / 5
• Subtask 12: 0.0833 / 5