| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 220092 | rama_pang | Minerals (JOI19_minerals) | C++14 | 72 ms | 3968 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "minerals.h"
#include <bits/stdc++.h>
using namespace std;
bool query(int x) {
static int last = -1;
int cur = Query(x);
bool res = (cur == last);
last = cur;
return res;
}
void Solve(vector<int> A, vector<int> B, bool isBFull) {
int n = A.size();
if (n == 0) return;
if (n == 1) return Answer(A[0], B[0]);
int piv = n / 2;
vector<int> Aleft, Aright;
vector<int> Bleft, Bright;
if (isBFull) { // All of B is in the box, so we remove the second half
for (int i = piv; i < n; i++) {
query(B[i]); // remove elements in B such that only the first half remains
}
} else { // None of B is in the box, so we add the first half
for (int i = 0; i < piv; i++) {
query(B[i]); // add elemens in B such that only the first half remains
}
}
for (int i = 0; i < n; i++) {
if (Aright.size() == n - piv) {
Aleft.emplace_back(A[i]);
} else if (Aleft.size() == piv) {
Aright.emplace_back(A[i]);
} else if (query(A[i])) {
Aleft.emplace_back(A[i]); // A[i] has a match is in the first half of B
} else {
Aright.emplace_back(A[i]); // A[i] has a match in the second half of B
}
}
for (int i = 0; i < piv; i++) Bleft.emplace_back(B[i]);
for (int i = piv; i < n; i++) Bright.emplace_back(B[i]);
return Solve(Aleft, Bleft, true), Solve(Aright, Bright, false);
}
void Solve(int N) {
vector<int> A, B;
for (int i = 1; i <= 2 * N; i++) {
if (query(i)) { // the current element is already in the box
B.emplace_back(i);
} else { // the current element is a new element
A.emplace_back(i);
}
}
return Solve(A, B, true);
}
컴파일 시 표준 에러 (stderr) 메시지
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