[通用算法] 求30年的5-10月的逐日降水平均值

Program main
Integer:: nx=240, ny=121,nt=31,i,j,m,k
real surp(240,121,31,30),dp(240,121,31)
character::  filename, year

do k=1981,2010  
   write(year,'(i4)') k
   filename=year//'.bin'
   open(11,file='F:\cut\surp-\'//filename,form='binary')
   do m=1,nt
     do j=1,ny
	   do i= 1,nx
       read(12)surp(i,j,m,k-1980)
	  enddo							
 	 enddo
   enddo
   Close(11)
enddo


do i= 1,nx
do j= 1,ny
do m= 1 ,nt
   ss=0.0
   do k = 1,30
      ss= ss+surp(i,j,m,k)
   enddo
   dp(i,j,m)=ss/30.0
enddo
enddo
enddo
open(12,file='F:\surp.interim.daily.climato.bin',form='binary')
write(12)(((dp(i,j,m),i=1,240),j=1,121),m=1,31)	
End Program	main

open(11,file='F:\ERA_Interim_1.5\cut5-10\surp.interim.daily.climato.bin',FORM='UNFORMATTED',ACCESS='DIRECT',action='read',RECL=NX*121)
LOOPDAY: DO K=1,nt
  print *,'Begin time = ',k
        HT  =-9.99E33        
        ffG =-9.99e33       
        tg  =-9.99e33        
        ug  =-9.99e33        
        wg  =-9.99e33        
        ft  =-9.99e33        
        fv  =-9.99e33        
        fw  =-9.99e33       
  read (11,rec=k)((orog1(i,j),i=1,nx),j=1,121)
  pres(:,1:119)=orog1(:,2:120)
  if (yrev) then
    pres(:,1:119)=pres(:,119:1:-1)
  endif