Originally posted by snoopyzhao at 2011-05-02 11:13:46:
下载不下来，呵呵……

如果你知道矩阵的行数或列数，自己计算一下其中的数据个数，然后把它们全部读到一个一维数组中，或者直接到二维数组也可以……

类似于这样……

read(*,*) (a(i), i = 1, 100)

...

program overlap

real, dimension(88,88) :: ol
character(len=80) :: line,lfm
integer :: i, j, k, ios
integer :: idat = 12

   open(unit=idat, file='overlap.txt')

   do i=1,88
      do j=1,88
         ol(i,j) = 0.0
      end do
   end do

   do i=1,88,5
      read(idat,'(a)', iostat = ios ) line
      if (ios < 0) exit
      read(idat,'(a)') line
      do j=i,88
         read(idat,'(a)') line
         write(lfm, '(a,i0,a)') '(tr1,',min(i+4,j),'(tr2,f9.6))'
         read(line,lfm) (ol(j,k),k=i,min(i+4,j))
!        write(*,lfm) (ol(j,k),k=i,min(i+4,j))  
      end do
      read(idat,'(a)', iostat = ios) line
      if (ios < 0) exit
   end do

end program overlap

program overlap

real, dimension(88,88) :: ol
character(len=80) :: line,lfm
integer :: i, j, k, l, ios
integer :: idat = 12

   open(unit=idat, file='overlap.txt')

   do i=1,88
      do j=1,88
         ol(i,j) = 0.0
      end do
   end do

   do i=1,88,5
      read(idat,'(a)', iostat = ios ) line
      if (ios < 0) exit
      read(idat,'(a)') line
      l=0
      do j=i,88
         if (l < 5) l=l+1
         read(idat,'(a)') line
         write(lfm, '(a,i0,a)') '(tr1,',l,'(tr2,f9.6))'
!        print *, lfm, min(i+4,j)
         read(line,lfm) (ol(j,k),k=i,min(i+4,j))
!        write(*,lfm) (ol(j,k),k=i,min(i+4,j))  
      end do
      read(idat,'(a)', iostat = ios) line
      if (ios < 0) exit
   end do

end program overlap

Originally posted by snoopyzhao at 2011-05-02 15:42:17:
刚才那个里面关于 lfm 的赋值有点儿问题，尽管不影响使用，呵呵……下面这个是改正后的。

[code]
program overlap

real, dimension(88,88) :: ol
character(len=80) :: line,lfm
integer :: i, j, k, l, ...

Originally posted by kathy2008 at 2011-05-02 15:57:54:
继续请教: idat = 12，是怎么样
确定把12付给idat这个变量的呢？也就是说为什么要选择12，而不是其他数字，谢谢。