tipsv/formpi.f90 at master · kensuke1984/tipsv · GitHub

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!cccccccccc       these subroutines are for mpi
!ccc
subroutine simplesplit(istart, iend, n, imin, imax)
!c     this routine returns the imin(i) and imax(i) (i=1,...,n)
!c      separates istart--iend into n parts
!c     each part contains imin(i) -- imax(i) i= 1, ..., n
!c
!c     iend-istart+1 = inum = n * deltai + remainder
!c     remainder is included in i th irange
!c     iend = n*deltai + remainder + istart -1
!c     istart = iend-remainder-n*deltai+1
!c     n * deltai = iend-(start+remainder) +1
!c
!c     remain :: istart, istart+1, ..., (istart + remainder-1)
!c     the others ::
!c     i = (istart+iamari), ...., imax ------------- (deltai*n)
!c     i(i) = istart+remainder+(i-1)*deltai,
!c     ......istart+remainder+n*deltai-1
!cccccccccccccccccccccccccccccccccccccc
!c     input parameters
    implicit none
    integer, intent(in) :: istart, iend
    integer, intent(in) :: n  ! the number of processors
    !c
    integer, dimension(*), intent(out) :: imin, imax
    !c
    integer :: remainder
    integer :: inum
    integer :: deltai
    !c
    integer :: i
    !c
    inum = iend-istart+1
    remainder = mod (inum, n)
    deltai = (inum-remainder)/n
    !c
    imin (1) = istart
    imax (1) = istart + remainder-1 + deltai
    do i = 2, n
        imin(i)= istart + remainder + (i-1) * deltai
        imax(i)= istart + remainder + i * deltai -1
    enddo
    return
end

subroutine trianglesplit (istart, iend, n, imin, imax)
!c     this routine returns the imin(i) and imax(i) (i=1,...,n)
!c      separates istart--iend into n parts
!c     each part contains imin(i) -- imax(i) i= 1, ..., n
!c
!c     iend-istart+1 = inum
!c
!c     Assume that the cpu time t for i th omega is a * i
!c     then we divide a* 0.5 * i **2 into n parts.
!c
!c     return imin imax which satisfy above assumption
!c     input parameters
    implicit none
    integer, intent(in) :: istart, iend
    integer, intent(in) :: n  ! the number of processors
    !c
    integer, dimension(*), intent(out) :: imin, imax
    !c
    integer :: i,inum
    integer:: x(0:n)
    double precision :: s  !  0.5 *  iend **2 / n
    !c
    inum = iend-istart+1
    s =  iend *iend / n
    !c
    x(0) = istart
    do i = 1, n
        x(i)= s+x(i-1)**2
        x(i) = x(i)**0.5
    enddo


    imin(1:n)= x(0:n-1)
    imax(1:n)= x(1:n)-1

    imax (n) = iend
    return

end