DIMENSION A(2,3) A=175 A(2,2)=2*A(2,2) A(5),A(1,vf程序设计

DIMENSION A(2,3) A=175 A(2,2)=2*A(2,2) A(5),A(1,vf程序设计 vf中do while-enddo循环的嵌套怎么执行循环的?急set talk offclear dimension a(2,3)i=1do while i 麻烦帮我详细讲解vf中的题目：DIMENSION A(2,3) A=175 A(2,2)=2*A(2,2) A(5),A(1,2) 输出为：350 175 dimension a (3) s=0 for k=1 to 3 a(k)=3-k s=s+k endfor a(1),a(2),a(3),s VFP中 CLEAR DIMENSION a（3） FOR N=1 TO 3 S=4 a(N)=N*2+1 S=S+a(N) ENDFOR S 关于vf的题,DIMENSION A(6)FOR K=1 TO 6A(K)=20-2*kENDFORK=5DO WHILE K>=1A(K)=A(K)-A(K+1)K=K-1ENDDOA(1),A(3),A(5) Fortran里面dimension A(6,3)是什么意思?dimension是定义一个一维数组吗?那么A（6,3）应该怎么解释呢,谢谢各位~~~ a new dimension of life a new dimension of life 急用!11 a(1,:)=perms([1 2 3 4 5])在matlab中为什么出错为Subscripted assignment dimension mismatchfor r=1:1:120a(1,:)=perms([1 2 3 4 5 ])以上是否可以输出矩阵a第一行的所有情况 我有一个复数数组,我想用fortran的write输出成6*6矩阵,format里写什么?parameter(n=6)dimension a(n,n)complex ::aa(1,1)=(1.0,0.0)a(1,2)=(1.0,2.0)a(1,3)=(2.0,1.0)a(1,4)=(0.0,0.0)a(1,5)=(3.0,0.0)a(1,6)=(2.0,0.0)a(2,1)=(1.0,-2.0)a(2,2)=(2 fortran求找错program dsdimplicit nonereal(8) ::a(2,2)data a/1,3,2,4/call brinv(a,2)print*,aend program dsdsubroutine brinv(a,n)dimension a(n,n),is(n),js(n)double precision a,t,dl=1do 100 k=1,nd=0.0do 10 i=k,ndo 10 j=k,nif(abs(a(i,j)).gt.d) thend=a 请问这道题的答案是选A还是C下列程序运行结果是 Dimension x(6) Store 1 to x(1),x(2) For i=3 to 5 X(i)=2*x(i-1) Endfor x(6) 选择A.F B.T C.16 D.32 求fortran语言改正!将10个数的数组A中最大的元素与A(10)互换program mainimplicit nonereal,dimension(10)::areal term,maxinteger iread*,amax=a(1)DO I=2,10IF(a(i)>=max)thenmax=a(i)term=maxmax=a(10)a(10)=termend ifend doprint*,aend有什 谁能把每句fortran语言都给我翻译一下integer row,i,j,k,nreal sreal,dimension(:,:),allocatable::a,l,ureal,dimension(:),allocatable::b,x,yread*,rowallocate(a(row,row))read*,((a(i,j),j=1,row),i=1,row)allocate(b(row))read*,(b(i),i=1,row)alloc vfp中 dimension a (3,4) a=8 a(3,4)=5 a(3) 输出结果是 a(3)=8 请问为什么结果是8,怎么算出来的. fortran语言 DATA数组赋值问题 DIMENSION A（4,3） DATA A/3*-1.0,3*-2.0,3*-3.0,3*-4.0/请教A（1,1）A（1,2）A（1,3）这些数组的值赋给的是多少? matlab 老说我Subscripted assignment dimension mismatch.这是我的代码a=[0,2,2,0,-3,0];l=length(a);w=(0:0.1:100);p=length(w);yw=zeros(p,l-1);for m=1:l-1 for n=1:p if m == 1 yw(n,m)=((a(m)*(1-exp(-1i*w(n)*0.5))/... (1i*w( dimension a(10)for i=1 to10a（i）=i+1endforfor i=1 to 9for j=i+1 to 10if a(i)