fortran编了一个两数组相加求和，再求输出数组的最大值的程序，怎么算出来的数字有点问题

嗯，你的 ld 都没有定义，也没有赋值。就这一点你的程序就完全没法继续看了。

Fortran 的优势就是矩阵运算。你的程序可以非常简单。

program main

real a(3,3),b(3,3),c(3,3),cmax

integer n,m

print*,'输入m'

read*,m

print*,'输入n'

read*,n

print*,'输入矩阵a'

read*,a

print*,'输入b'

read*,b

call smv(a,b,c,m,m,n,cmax)

print*,'matrice c',c,'cmax',cmax

end

subroutine smv(a,b,c,ld,m,n,cmax)

real a(ld,m),b(ld,m),c(ld,m)

integer n,m

c = a + b

cmax = maxval(c)

end subroutine

这两句代码就可以了。

我建议你看一下这个文章，关于定义数组和变量的一些常识：fcode.cn/guide-43-1.html

不保证全对 你在检查一下吧

program BIN_SUM

implicit none

integer:: x,y,z,counter

integer::array_x(8),array_y(8),array_sum(8),array_carry(8)

!!initialize all the arrays to zero

do counter=1,8

array_x(counter)=0

array_y(counter)=0

array_sum(counter)=0

array_carry(counter)=0

end do

write(*,*) "please enter two positive number X and Y which are smaller than 256"

read(*,*) x

read(*,*) y

z=x+y

if (z>256) then

write(*,*) "the sum of two number must be smaller than 256"

end if

if(x>=0 .and. x<=256 .and. y>=0 .and. y<=256) then

!divide x by 2, and put the remainder into each colunm of the array

do counter=8,1,-1

array_x(counter)=mod(x,2)

x=x/2

end do

write(*,*)"your X number in binary form is:"

write(*,*) (array_x(counter),counter=1,8)

do counter=8,1,-1

array_y(counter)=mod(y,2)

y=y/2

end do

write(*,*)"your Y number in binary form is :"

write(*,*) (array_y(counter),counter=1,8)

else

write(*,*) "wrong walue! please type your number between 0 and 256"

end if

!!use do loop to perform the calculation, if array_x + array _y = 2 put 1 in the carry array and put zero in the sum array

do counter=8,1,-1

if ((array_x(counter)+array_y(counter)+array_carry(counter))==1)then

array_sum(counter)=1

array_carry(counter-1)=0

end if

if ((array_x(counter)+array_y(counter)+array_carry(counter))==2) then

array_sum(counter)=0

array_carry(counter-1)=1

end if

if ((array_x(counter)+array_y(counter)+array_carry(counter))==3) then

array_sum(counter)=1

array_carry(counter-1)=1

end if

end do

write(*,*)"the sum of two numbers is :",z

write(*,*)"the sum of two number in binary form is :"

write(*,*) (array_sum(counter),counter=1,8)

end program BIN_SUM

“出现则输出相应的第二列第三列，进行加减运算”

这句里不清楚是否需要输出第一列的数据，我就只输出了第二、第三列。

需要第一列数据的话，改一下write21。

还有加减运算不清楚具体需要对哪些数据进行加法还是减法运算，

我就只对第二列和第三列数据进行了求和。

需要其他加减结果的话，改一下write22。

      !

      implicit none

      integer a(10,3),b(5),i,j

      ! 读取文件A

      open(11,file='A.txt')

      do i=1,10

        read(11,*) (a(i,j),j=1,3)

      end do

      close(11)

      ! 读取文件B

      open(12,file='B.txt')

      do i=1,5

        read(12,*) b(i)

      end do

      close(12)

      

      ! 比较A和B，并输出结果

      open(21,file='result.txt')

      open(22,file='sum.txt')

      do i=1,5

        do j=1,10

          if(a(j,1).eq.b(i)) then

            write(21,*) a(j,2),a(j,3)

            write(22,*) a(j,2)+a(j,3)

          end if

        end do

      end do

      close(21)

      close(22)

      stop

      end

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