제출 #1262439

#제출 시각아이디문제언어결과실행 시간메모리
1262439Canuc80kTriple Peaks (IOI25_triples)C++20
25 / 100
174 ms84872 KiB
#include "bits/stdc++.h" using namespace std; using ll = long long; using ld = long double; using ull = unsigned long long; ll res = 0; void add(ll i, ll j, ll k) {res ++;} ll countr(const vector<int> &a, const vector<int> &b) { int n = (int)a.size(); int SQ = sqrt(n) + 1; ll res = 0; // Lưu vị trí các giá trị trong a và b unordered_map<int, vector<int>> posA, posB; for (int i = 0; i < n; i++) { posA[a[i]].push_back(i); posB[b[i]].push_back(i); } // Xác định heavy value trong a vector<int> heavy; for (auto &p : posA) { if ((int)p.second.size() > SQ) heavy.push_back(p.first); } // Tiền xử lý cho heavy: cnt[valA][valB] = số cặp (i) với a[i]=valA, b[i]=valB unordered_map<int, unordered_map<int,int>> cnt; for (int hv : heavy) { for (int idx : posA[hv]) { cnt[hv][b[idx]]++; } } // Đếm số j thỏa điều kiện prefix để xử lý light unordered_map<int, unordered_map<int,int>> seen; // seen[valA][valB] = số j đã đi qua với a[j]=valA, b[j]=valB for (int k = 0; k < n; k++) { int x = a[k]; // cần a[j] = a[k] = x int y = b[k]; // cần a[i] = b[k] = y // Heavy: dùng cnt precompute if (posA[x].size() > SQ) { if (cnt[x].count(y)) res += cnt[x][y]; } // Light: duyệt hết các i ứng với a[i] = y else { for (int i : posA[y]) if (i < k) { int val = b[i]; if (seen[x].count(val)) res += seen[x][val]; } } // update seen cho j = k seen[a[k]][b[k]]++; } return res; } long long count_triples(std::vector<int> H) { // #TH: a[k] max for (int k = 2; k < H.size(); k ++) { int i = k - H[k]; if (i < 0 || H[i] >= H[k]) continue; int j1 = k - H[i], j2 = i + H[i]; if (j1 < 0 || H[j1] >= H[k] || H[j1] != j1 - i) j1 = -1; if (j2 >= k || H[j2] >= H[k] || H[j2] != k - j2) j2 = -1; if (j1 != -1) add(i, j1, k); if (j2 != -1 && j1 != j2) add(i, j2, k); } // #TH: a[i] max for (int i = 0; i + 2 < H.size(); i ++) { int k = i + H[i]; if (k >= H.size() || H[k] >= H[i]) continue; int j1 = k - H[k], j2 = i + H[k]; if (j1 < 0 || H[j1] >= H[i] || H[j1] != j1 - i) j1 = -1; if (j2 >= k || H[j2] >= H[i] || H[j2] != k - j2) j2 = -1; if (j1 != -1) add(i, j1, k); if (j2 != -1 && j1 != j2) add(i, j2, k); } // #TH: a[j] max, j - i = a[i] for (int i = 0; i + 2 < H.size(); i ++) { int j = i + H[i]; if (j >= H.size() || H[j] <= H[i]) continue; int k = i + H[j]; if (k >= H.size() || H[k] >= H[j] || k - H[k] != j || k - j == H[i]) k = -1; if (k != -1) add(i, j, k); } // #TH: a[j] max, j - i = a[k] // for (int i = 0; i + 2 < H.size(); i ++) { // for (int k = i + H[i] + 1; k < H.size(); k ++) { // int j = i + H[k]; if (j != k - H[i] || j >= H.size()) continue; // if (H[i] >= H[j] || H[j] <= H[k] || k - i != H[j] || j >= k) continue; // // cout << "OKE: " << i << ' ' << j << ' ' << k << endl; // add(i, j, k); // } // } vector<int> a, b; for (int i = 0; i < H.size(); i ++) a.push_back(H[i] + i); for (int i = 0; i < H.size(); i ++) b.push_back(i - H[i]); // for (int i = 0; i < a.size(); i ++) // for (int j = i + 1; j < a.size(); j ++) // for (int k = j + 1; k < a.size(); k ++) // if (a[i] == b[k] && a[j] == a[k] && b[i] == b[j]) res ++; res += countr(a, b); return res; } std::vector<int> construct_range(int M, int K) { // vector<int> res; res.push_back(1); // for (int i = 1; i < M; i ++) res.push_back(i); // return res; }

컴파일 시 표준 에러 (stderr) 메시지

triples.cpp: In function 'std::vector<int> construct_range(int, int)':
triples.cpp:115:1: warning: no return statement in function returning non-void [-Wreturn-type]
  115 | }
      | ^
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