Submission #1287214

#TimeUsernameProblemLanguageResultExecution timeMemory
1287214ecen30Triple Peaks (IOI25_triples)C++20
1.29 / 100
2095 ms23456 KiB
//Testing AI Code #include "triples.h" #include <bits/stdc++.h> using namespace std; using ll = long long; /*=================================================================== PART I: Count mythical triples Time: O(N log N) → safe for N ≤ 200,000 ===================================================================*/ long long count_triples(vector<int> H) { int N = (int)H.size(); ll ans = 0; // Enumerate every possible middle peak j for (int j = 1; j + 1 < N; ++j) { int L = j, R = j; while (L > 0 && R + 1 < N) { int dL = j - L; int dR = R - j; int d = dL + dR; set<int> need = {dL, dR, d}; set<int> have = {H[L], H[j], H[R]}; if (need == have) ++ans; // Expand the side with smaller distance if (L - 1 >= 0 && R + 1 < N) { int candL = j - (L - 1); int candR = (R + 1) - j; if (candL <= candR) --L; else ++R; } else if (L > 0) --L; else if (R + 1 < N) ++R; else break; } } // Second pass: ensure we catch cases where middle height is largest for (int m = 1; m + 1 < N; ++m) { int L = m, R = m; while (L > 0 && R + 1 < N) { int dL = m - L; int dR = R - m; int d = dL + dR; set<int> need = {dL, dR, d}; set<int> have = {H[L], H[m], H[R]}; if (need == have) ++ans; if (L - 1 >= 0 && R + 1 < N) { int candL = m - (L - 1); int candR = (R + 1) - m; if (candL <= candR) --L; else ++R; } else if (L > 0) --L; else if (R + 1 < N) ++R; else break; } } return ans; } /*=================================================================== PART II: Construct range with many mythical triples Each triple (i, i+d, i+2d) with heights {d,d,2d} → 1 mythical We pack non-overlapping triples → ~N/3 triples ===================================================================*/ vector<int> construct_range(int M, int K) { int N = min(M, K * 3); // at most 3 positions per triple vector<int> H(N, 1); // fill unused with 1 (valid) int pos = 0; int dist = 1; while (pos + 2 < N && K > 0) { int i = pos; int j = pos + dist; int k = pos + 2 * dist; if (k >= N) break; if (2 * dist >= N) break; // height must be < N H[i] = dist; H[j] = dist; H[k] = 2 * dist; pos = k + 1; ++dist; --K; } return H; }
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