結果
問題 |
No.3180 angles sum
|
ユーザー |
![]() |
提出日時 | 2025-06-13 22:40:19 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 2,106 bytes |
コンパイル時間 | 268 ms |
コンパイル使用メモリ | 82,372 KB |
実行使用メモリ | 78,476 KB |
最終ジャッジ日時 | 2025-06-14 01:43:19 |
合計ジャッジ時間 | 9,178 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 16 WA * 1 |
ソースコード
class Vec2: def __init__(self, x, y): self.x = x self.y = y def __repr__(self): return f'({self.x}, {self.y})' def dot(self, o): return self.x * o.x + self.y * o.y def cross(self, o): return self.x * o.y - self.y * o.x def normalized(self): m = self.mag() if m != 0 and m != 1: return Vec2(self.x / m, self.y / m) return self def magsq(self): return self.x * self.x + self.y * self.y def mag(self): return sqrt(self.magsq()) def rotate(self, rad): x = self.x * cos(rad) - self.y * sin(rad) y = self.x * sin(rad) + self.y * cos(rad) return Vec2(x, y) def rot90ccw(self): return Vec2(-self.y, self.x) def distance(self, other) -> float: return (self - other).mag() def __eq__(self, other): return self.x == other.x and self.y == other.y def __ne__(self, other): return not self.__eq__(other) def __lt__(self, other): if self.x == other.x: return self.y < other.y return self.x < other.x def __add__(self, other): return Vec2(self.x + other.x, self.y + other.y) def __sub__(self, other): return Vec2(self.x - other.x, self.y - other.y) def __mul__(self, d): return Vec2(self.x * d, self.y * d) def __rmul__(self, d): return self.__mul__(d) def __truediv__(self, d): return Vec2(self.x / d, self.y / d) def __neg__(self): return Vec2(-self.x, -self.y) def __iadd__(self, other): self.x += other.x self.y += other.y return self def __isub__(self, other): self.x -= other.x self.y -= other.y return self def solve(): AX, AY, BX, BY, CX, CY = map(int, input().split()) a = complex(AX, AY) b = complex(BX, BY) ab = a * b s = Vec2(ab.real, ab.imag) t = Vec2(CX, CY) return s.cross(t) == 0 T = int(input()) for _ in range(T): ans = solve() if ans: print('Yes') else: print('No')