結果

問題 No.2953 Maximum Right Triangle
ユーザー osada-yum
提出日時 2024-11-16 01:02:53
言語 Fortran
(gFortran 14.2.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,661 bytes
コンパイル時間 1,611 ms
コンパイル使用メモリ 32,256 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-11-16 01:02:56
合計ジャッジ時間 1,170 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 6
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 #
プレゼンテーションモードにする

!> This file was processed by `fypp`.
!> Today's fortune: "Lucky:)", really OK?
!> '' .
program f902953
use, intrinsic :: iso_fortran_env
!> auto use module
implicit none
integer(int32) :: t
integer(int32) :: i
read(input_unit, *) t
do i = 1, t
call solve()
end do
contains
impure subroutine solve()
integer(int64) :: d, x, y
integer(int64) :: g, nx, ny
read(input_unit, *) d, x, y
if (x == 0) then
write(output_unit, '(i0)') y * d
return
else if (y == 0) then
write(output_unit, '(i0)') x * d
return
end if
!> (nx, ny) == (y, -x) /g (x, y) .
g = gcd(x, y)
nx = y / g
ny = - x / g
block
!> (x, y)
!> nx >= 0.
!> x + t * nx <= d.
integer(int64) :: ans
integer(int64) :: bx, by
integer(int64) :: n_dist_sq
n_dist_sq = nx ** 2 + ny ** 2
ans = 0_int64
if (nx < 0) then
nx = - nx
ny = - ny
end if
associate(t => (d - x) / nx)
!> x + t * nx <= d
if (t > 0) then
bx = x + t * nx
by = y + t * ny
if (0 <= by .and. by <= d) then
ans = max(ans, t * g * n_dist_sq)
end if
end if
end associate
associate(t => x / nx)
!> x - t * nx >= 0
if (t > 0) then
bx = x - t * nx
by = y - t * ny
if (0 <= by .and. by <= d) then
ans = max(ans, t * g * n_dist_sq)
end if
end if
end associate
if (ny < 0) then
nx = - nx
ny = - ny
end if
associate(t => (d - y) / ny)
!> y + t * ny <= d
if (t > 0) then
bx = x + t * nx
by = y + t * ny
if (0 <= bx .and. bx <= d) then
ans = max(ans, t * g * n_dist_sq)
end if
end if
end associate
associate(t => y / ny)
!> y - t * ny >= 0
if (t > 0) then
bx = x - t * nx
by = y - t * ny
if (0 <= bx .and. bx <= d) then
ans = max(ans, t * g * n_dist_sq)
end if
end if
end associate
write(output_unit, '(i0)') ans
end block
end subroutine solve
pure integer(int64) function gcd(a, b) result(res)
integer(int64), intent(in) :: a, b
integer(int64) :: arr(1:2)
arr(1:2) = [max(abs(a), abs(b)), min(abs(a), abs(b))]
do while (arr(2) /= 0)
arr(1:2) = [arr(2), mod(arr(1), arr(2))]
end do
res = arr(1)
end function gcd
end program f902953
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0