結果
問題 |
No.3115 One Power One Kill
|
ユーザー |
|
提出日時 | 2025-04-19 11:57:28 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,603 bytes |
コンパイル時間 | 2,027 ms |
コンパイル使用メモリ | 193,312 KB |
実行使用メモリ | 26,252 KB |
平均クエリ数 | 22.00 |
最終ジャッジ日時 | 2025-04-19 11:57:34 |
合計ジャッジ時間 | 5,884 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | WA * 20 |
ソースコード
#include <bits/stdc++.h> using namespace std; const int MOD = 1e9+7; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); // Initial guess long long A = 2; // initial A guess long long B = 1000; // initial B guess // Output the initial A and B cout << A << " " << B << "\n" << flush; // Read the response K long long K; cin >> K; // Read the current guessed X' long long X_prime; cin >> X_prime; // Read the result code int ret; cin >> ret; // For demonstration: // A simple loop to adjust guesses based on previous K. // In a real scenario, you might implement a more intelligent method. for (int attempt = 0; attempt < 20; attempt++) { // Based on previous K, attempt to narrow down X. // For simplicity, keep A fixed and iterate B. B += 1000; // increase B for new attempt; in practice, choose adaptively cout << A << " " << B << "\n" << flush; cin >> K; cin >> X_prime; cin >> ret; if (ret == 1) { // Guess is correct, can terminate break; } // Otherwise, continue guessing... // Implement further logic here to refine guesses based on K // For example, if K divides some candidate values, narrow down. } // After guesses, output your final guess for X' // (or, in a real scenario, attempt to derive X from collected info) // Here, just output the last X' obtained cout << X_prime << "\n" << flush; }