結果
| 問題 |
No.3223 K-XOR Increasing Sequence
|
| コンテスト | |
| ユーザー |
woshinailong
|
| 提出日時 | 2025-08-01 22:57:36 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,756 bytes |
| コンパイル時間 | 733 ms |
| コンパイル使用メモリ | 79,232 KB |
| 実行使用メモリ | 12,972 KB |
| 最終ジャッジ日時 | 2025-08-01 22:57:53 |
| 合計ジャッジ時間 | 16,079 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | WA * 3 |
| other | WA * 70 |
ソースコード
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int N, K;
cin >> N >> K;
vector<int> A(N);
for (int i = 0; i < N; ++i) {
cin >> A[i];
}
// Separate the numbers into even and odd
vector<int> even, odd;
for (int num : A) {
if (num % 2 == 0) {
even.push_back(num);
} else {
odd.push_back(num);
}
}
// Check if it's possible to form a subsequence of K elements with an even sum
if (K == 0) {
cout << "Yes" << endl;
return 0; // Empty subsequence
}
// We need to have an even sum
// Case 1: All numbers in subsequence are even
if (even.size() >= K) {
cout << "Yes" << endl;
for (int i = 0; i < K; ++i) {
cout << even[i] << " ";
}
cout << endl;
return 0;
}
// Case 2: Use odd numbers
int oddCount = odd.size();
// We need an even count of odd numbers to keep the total sum even
// (0, 2, 4, ...) odd numbers can be used
int requiredOdds = (K - even.size()) % 2; // This should be even to maintain the parity
// Check if we can satisfy the count
if (even.size() + oddCount >= K && oddCount >= requiredOdds) {
cout << "Yes" << endl;
for (int i = 0; i < even.size(); ++i) {
cout << even[i] << " ";
}
// Need to add odd numbers now
int oddsToTake = K - even.size();
if (oddsToTake % 2 != 0) {
oddsToTake--; // Make it even
}
for (int i = 0; i < oddsToTake; ++i) {
cout << odd[i] << " ";
}
cout << endl;
return 0;
}
cout << "No" << endl;
return 0;
}
woshinailong