結果

問題 No.132 点と平面との距離
ユーザー Mao-beta
提出日時 2024-03-15 11:40:46
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,382 ms / 5,000 ms
コード長 1,497 bytes
コンパイル時間 172 ms
コンパイル使用メモリ 82,424 KB
実行使用メモリ 77,844 KB
最終ジャッジ日時 2024-09-30 00:00:06
合計ジャッジ時間 2,794 ms
ジャッジサーバーID
(参考情報)
judge1 / judge2
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
other AC * 3
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ソースコード

diff #
プレゼンテーションモードにする

import sys
import math
import bisect
from heapq import heapify, heappop, heappush
from collections import deque, defaultdict, Counter
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
sys.setrecursionlimit(1000000)
MOD = 10 ** 9 + 7
MOD99 = 998244353
input = lambda: sys.stdin.readline().strip()
NI = lambda: int(input())
NMI = lambda: map(int, input().split())
NLI = lambda: list(NMI())
SI = lambda: input()
SMI = lambda: input().split()
SLI = lambda: list(SMI())
EI = lambda m: [NLI() for _ in range(m)]
def cross3(X, Y):
return [X[(i+1)%3]*Y[(i+2)%3]-X[(i+2)%3]*Y[(i+1)%3] for i in range(3)]
def dist_P_from_plane_of_3points(Qs):
Q01 = [Qs[1][i] - Qs[0][i] for i in range(3)]
Q02 = [Qs[2][i] - Qs[0][i] for i in range(3)]
R = cross3(Q01, Q02)
a, b, c = R
x, y, z = Qs[0]
d = a*x + b*y + c*z
return abs(d / math.sqrt(a**2+b**2+c**2))
def main():
N = NI()
# P = [int(s.replace(".", "")) for s in SLI()]
P = list(map(float, SMI()))
XYZ = []
for _ in range(N):
# xyz = [int(s.replace(".", "")) for s in SLI()]
xyz = list(map(float, SMI()))
xyz = [p-x for p, x in zip(P, xyz)]
XYZ.append(xyz)
ans = 0
for i in range(N):
for j in range(i+1, N):
for k in range(j+1, N):
ans += dist_P_from_plane_of_3points([XYZ[i], XYZ[j], XYZ[k]])
# print(ans / 10**6)
print(ans)
if __name__ == "__main__":
main()
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