제출 #596935

#제출 시각아이디문제언어결과실행 시간메모리
596935jophyyjhDetecting Molecules (IOI16_molecules)C++14
69 / 100
39 ms5676 KiB
/** * If we've got a subset S and two elements x, y (not in S). We must not have: * sum(S + {x}) > u, sum(S + {y}) < v, * As this immediately violates the given constraint. We now can propose a solution. * * Time Complexity: O(n * log(n)) * Implementation 1 */ #include <bits/stdc++.h> #include "molecules.h" struct val_t { int pos, val; }; std::vector<int> find_subset(int l, int u, std::vector<int> w) { int n = w.size(); std::vector<val_t> values(n); for (int k = 0; k < n; k++) values[k].val = w[k], values[k].pos = k; std::sort(values.begin(), values.end(), [](const val_t& v1, const val_t& v2) { return v1.val < v2.val; }); std::vector<int> prefix_sum(n + 1); prefix_sum[0] = 0; for (int i = 0; i < n; i++) prefix_sum[i + 1] = prefix_sum[i] + values[i].val; bool found = false; std::vector<int> ans; for (int i = n; i >= 1 && !found; i--) { int j = i; for (int step = n / 2 + 1; step >= 1; step /= 2) { while (j - step >= 0 && prefix_sum[i] - prefix_sum[j - step] <= u) j -= step; } if (prefix_sum[i] - prefix_sum[j] >= l) { assert(prefix_sum[i] - prefix_sum[j] <= u); found = true; for (int k = j; k < i; k++) ans.emplace_back(k); } else if (j > 0) { if (values[0].val + prefix_sum[i] - prefix_sum[j] <= u) { found = true; ans.emplace_back(0); for (int k = j; k < i; k++) ans.emplace_back(k); } } } if (found) { // test-driven development int sum = 0; for (int& a : ans) { sum += values[a].val; a = values[a].pos; } assert(sum >= l && sum <= u); return ans; } else { return std::vector<int>(); } }
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