結果

問題 No.2953 Maximum Right Triangle
ユーザー osada-yum
提出日時 2024-11-16 00:35:35
言語 Fortran
(gFortran 14.2.0)
結果
WA  
実行時間 -
コード長 2,609 bytes
コンパイル時間 1,601 ms
コンパイル使用メモリ 32,512 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-11-16 00:35:37
合計ジャッジ時間 1,949 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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ファイルパターン 結果
sample AC * 1
other AC * 2 WA * 4
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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

!> This file was processed by `fypp`.
!> Today's fortune: "Bad WA", really OK?
!> ランダムウォーク猿「'Suffix Array' で はっぴー.」
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 ベクトルは (x, y) ベクトルに垂直.
    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
      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 * (x ** 2 + y ** 2))
           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 * (x ** 2 + y ** 2))
           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 * (x ** 2 + y ** 2))
           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 * (x ** 2 + y ** 2))
           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
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