結果
問題 | No.622 点と三角柱の内外判定 |
ユーザー |
👑 |
提出日時 | 2019-05-10 00:01:42 |
言語 | Lua (LuaJit 2.1.1734355927) |
結果 |
AC
|
実行時間 | 2 ms / 1,500 ms |
コード長 | 1,709 bytes |
コンパイル時間 | 32 ms |
コンパイル使用メモリ | 6,684 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-07-02 00:53:28 |
合計ジャッジ時間 | 991 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 32 |
ソースコード
local x, y, z = {}, {}, {}for i = 1, 4 dox[i], y[i], z[i] = io.read("*n", "*n", "*n")endfor i = 2, 4 dox[i], y[i], z[i] = x[i] - x[1], y[i] - y[1], z[i] - z[1]endlocal e1sz = math.sqrt(x[2] * x[2] + y[2] * y[2] + z[2] * z[2])local e1 = {x[2] / e1sz, y[2] / e1sz, z[2] / e1sz}local innerpr = e1[1] * x[3] + e1[2] * y[3] + e1[3] * z[3]local e2 = {x[3] - innerpr * e1[1],y[3] - innerpr * e1[2],z[3] - innerpr * e1[3]}local e2sz = math.sqrt(e2[1] * e2[1] + e2[2] * e2[2] + e2[3] * e2[3])for i = 1, 3 do e2[i] = e2[i] / e2sz endlocal e3 = {x[4], y[4], z[4]}innerpr = e3[1] * e1[1] + e3[2] * e1[2] + e3[3] * e1[3]for i = 1, 3 doe3[i] = e3[i] - e1[i] * innerprendinnerpr = e3[1] * e2[1] + e3[2] * e2[2] + e3[3] * e2[3]for i = 1, 3 doe3[i] = e3[i] - e2[i] * innerprendlocal e3sz = math.sqrt(e3[1] * e3[1] + e3[2] * e3[2] + e3[3] * e3[3])for i = 1, 3 do e3[i] = e3[i] / e3sz endinnerpr = e3[1] * x[4] + e3[2] * y[4] + e3[3] * z[4]x[4] = x[4] - innerpr * e3[1]y[4] = y[4] - innerpr * e3[2]z[4] = z[4] - innerpr * e3[3]-- print("e1 " .. table.concat(e1, " "))-- print("e2 " .. table.concat(e2, " "))-- print("e3 " .. table.concat(e3, " "))-- print("v " .. x[4] .. " " .. y[4] .. " " .. z[4])--[[s(x2,y2,z2) + t(x3,y3,z3) = (x,y,z)x2 x3 s = xy2 y3 t y]]local epsilon = 1.0e-5local s, t = 0, 0local det = x[2] * y[3] - x[3] * y[2]if epsilon < math.abs(det) thens, t = (x[4] * y[3] - y[4] * x[3]) / det, (-x[4] * y[2] + y[4] * x[2]) / detelsedet = x[2] * z[3] - x[3] * z[2]s, t = (x[4] * z[3] - z[4] * x[3]) / det, (-x[4] * z[2] + z[4] * x[2]) / detend-- print(s, t)if(0 <= s and 0 <= t and s + t <= 1) then print("YES") else print("NO") end