結果
問題 |
No.3108 Luke or Bishop
|
ユーザー |
|
提出日時 | 2025-04-18 23:14:44 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 1,060 bytes |
コンパイル時間 | 886 ms |
コンパイル使用メモリ | 77,152 KB |
実行使用メモリ | 7,848 KB |
最終ジャッジ日時 | 2025-04-18 23:14:47 |
合計ジャッジ時間 | 1,862 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 26 |
ソースコード
#include <iostream> #include <cmath> using namespace std; int main() { long long Gx, Gy; cin >> Gx >> Gy; // Calculate minimum moves for Rook int rook_moves; if (Gx == 0 && Gy == 0) { rook_moves = 0; // Already at the goal } else if (Gx == 0 || Gy == 0) { rook_moves = 1; // Can reach in one move (horizontally or vertically) } else { rook_moves = 2; // Need two moves: one horizontal and one vertical } // Calculate minimum moves for Bishop int bishop_moves; if (Gx == 0 && Gy == 0) { bishop_moves = 0; // Already at the goal } else if (abs(Gx) == abs(Gy)) { bishop_moves = 1; // Can reach in one move (on a diagonal) } else { // For any point not on the main diagonals, we need 2 moves // With real numbers k, a bishop can reach any point in 2 moves bishop_moves = 2; } // Choose the piece that requires fewer moves int result = min(rook_moves, bishop_moves); cout << result << endl; return 0; }