95 lines
3.0 KiB
Fortran
95 lines
3.0 KiB
Fortran
! This code is supported by the website: https://www.guanjihuan.com
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! The newest version of this code is on the web page: https://www.guanjihuan.com/archives/45966
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module random_matrix_mod
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implicit none
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contains
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subroutine generate_random_matrix(n, A)
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integer, intent(in) :: n
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double precision, allocatable, intent(out) :: A(:,:)
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integer :: ierr
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allocate(A(n, n), stat=ierr)
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if (ierr /= 0) stop "内存分配失败"
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call init_random_seed()
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call random_number(A)
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end subroutine
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subroutine init_random_seed()
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integer :: i, n, clock, ierr
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integer, allocatable :: seed(:)
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call random_seed(size = n)
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allocate(seed(n), stat=ierr)
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if (ierr /= 0) stop "种子分配失败"
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call system_clock(count=clock)
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seed = clock + 37 * [(i - 1, i = 1, n)]
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call random_seed(put=seed)
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deallocate(seed)
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end subroutine
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end module
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program main
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use random_matrix_mod
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use f95_precision
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use blas95
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use lapack95
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implicit none
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integer, allocatable :: index1(:)
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integer n, i, j, info, ierr, stage, start, end_val, step, count_start, count_end, count_rate, test_0, test_times
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double precision, allocatable :: A(:,:)
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double precision time_used
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test_times = 20
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! 定义不同阶段的参数
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do stage = 1, 3
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select case(stage)
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case(1) ! 第一阶段：100-1000，步长100
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start = 100
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end_val = 1000
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step = 100
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case(2) ! 第二阶段：2000-10000，步长1000
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start = 2000
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end_val = 10000
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step = 1000
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case(3) ! 第三阶段：20000-30000，步长10000
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start = 20000
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end_val = 30000
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step = 10000
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end select
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n = start
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do while (n <= end_val)
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allocate(index1(n), stat=ierr)
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call system_clock(count_start, count_rate)
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test_0 = 1
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do while (test_0 <= test_times)
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call generate_random_matrix(n, A)
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call getrf(A, index1, info); call getri(A, index1, info) ! 使用 getrf 和 getri 对矩阵求逆。这时候 A 不再是原来的矩阵了，而是求逆后的矩阵。
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deallocate(A, stat=ierr)
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test_0 = test_0 + 1
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end do
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call system_clock(count_end)
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! 打印计算时间
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if (count_rate > 0) then
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time_used = real(count_end - count_start) / real(count_rate) / test_times
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write(*, '(a, I6, a, f12.6, a)') 'n = ', n, ' 的计算时间: ', time_used, ' 秒'
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else
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write(*,*) "无法获取计算时间"
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endif
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deallocate(index1, stat=ierr)
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n = n + step
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end do
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end do
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end program |