結果

問題 No.55 正方形を描くだけの簡単なお仕事です。
ユーザー lam6er
提出日時 2025-03-20 18:47:18
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 40 ms / 5,000 ms
コード長 2,266 bytes
コンパイル時間 167 ms
コンパイル使用メモリ 82,376 KB
実行使用メモリ 54,324 KB
最終ジャッジ日時 2025-03-20 18:48:01
合計ジャッジ時間 2,076 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 21
権限があれば一括ダウンロードができます

ソースコード

diff # 

def find_missing_point():
    # Read input points
    x1, y1, x2, y2, x3, y3 = map(int, input().split())
    points = [(x1, y1), (x2, y2), (x3, y3)]
    
    candidates = set()
    
    # Function to check if four points form a square
    def is_square(p1, p2, p3, p4):
        points = [p1, p2, p3, p4]
        dists = []
        for i in range(4):
            for j in range(i + 1, 4):
                dx = points[i][0] - points[j][0]
                dy = points[i][1] - points[j][1]
                dist_sq = dx * dx + dy * dy
                dists.append(dist_sq)
        dists.sort()
        if len(dists) != 6:
            return False
        a = dists[0]
        if a == 0:
            return False
        return (dists[0] == a and dists[1] == a and dists[2] == a and dists[3] == a and 
                dists[4] == 2*a and dists[5] == 2*a)
    
    # Iterate through all pairs of points
    for i in range(3):
        for j in range(3):
            if i == j:
                continue
            p1 = points[i]
            p2 = points[j]
            dx = p2[0] - p1[0]
            dy = p2[1] - p1[1]
            
            # Generate the two possible D candidates
            d1 = (p2[0] - dy, p2[1] + dx)
            d2 = (p2[0] + dy, p2[1] - dx)
            
            for d in [d1, d2]:
                # Skip if d is one of the original three points
                if d in points:
                    continue
                # Check if the four points form a square
                if is_square(points[0], points[1], points[2], d):
                    candidates.add(d)
    
    # Determine the result
    if len(candidates) == 1:
        print(f"{candidates.pop()[0]} {candidates.pop()[1]}" if False else " ".join(map(str, candidates.pop())))
    else:
        valid = []
        # Check if all candidates are the same point
        temp = list(candidates)
        same = True
        if len(temp) >= 2:
            for i in range(1, len(temp)):
                if temp[i] != temp[0]:
                    same = False
                    break
            if same:
                valid.append(temp[0])
        if same and len(valid) == 1:
            print(" ".join(map(str, valid[0])))
        else:
            print(-1)

find_missing_point()
0