[转帖]victory:fortran 语言实现

glise

victory:fortran 语言实现<BR>!用逆秩一方法求解<BR>!初值定义为单位矩阵，收敛速度比df(x)的逆慢<BR>!优点：若做其他题只需改动f的值<BR>!主函数<BR>implicit none<BR>integer i,count,n<BR>real(8),allocatable:: x(:,:),x0(:,:),f(:,:),s(:,:),b0(:,:),f2(:,:)<BR>real(8),allocatable:: k(:,:),y(:,:),f1(:,:),b1(:,:)<BR>print*,'请输入未知数的个数:'<BR>read*,n<BR>allocate(x(n,1),x0(n,1),f(n,1),s(n,1),b0(n,n),f2(n,1))<BR>allocate(k(n,n),y(n,1),f1(n,1),b1(n,n))<BR>x0=1.0;B0=0<BR>do i=1,n<BR>  b0(i,i)=1.0<BR>enddo<BR>count=0<BR>do while(count&lt;=30)<BR>call d(x0,f)                          !The first time use function_subroutine<BR>  f1=f<BR>  x=x0-matmul(b0,f) <BR>  s=x-x0<BR>  count=count+1<BR>  if(sum(abs(s))&lt;1e-5) exit         ! if the compute can't satisfy the precise,exit!<BR>  call d(x,f)<BR>  f2=f                                                 !The second time use function_subroutine<BR>  y=f2-f1<BR>  call g(s,b0,y,n,k)<BR>  b1=b0+k<BR>  x0=x<BR>  b0=b1<BR>enddo<BR>if(count&lt;=30)        then<BR>  do i=1,n<BR>    print*,x(i,1)<BR>  enddo<BR>  print*,'迭代次数为：',count<BR>else<BR>print*,'迭代发散'<BR>endif<BR>deallocate(x0,x,f,s,b0,f2,f1,k,y)<BR>end<BR><BR>!求解b1-b0的差值<BR>subroutine g(s,b,y,n,k)<BR>  implicit none<BR>  integer i,j,n<BR>  real(8)   c(1,1),m,s(n,1),b(n,n),y(n,1),k(n,n)<BR>  c=matmul(matmul(transpose(s),b),y)             !分母的值<BR>  m=c(1,1)<BR>  k=matmul(matmul(s-matmul(b,y),transpose(s)),b) !分子的值<BR>  do i=1,n<BR>    do j=1,n<BR>          k(i,j)=k(i,j)/m<BR>        enddo<BR>  enddo<BR>endsubroutine<BR><BR>!F(x)的定义<BR>subroutine d(x,f)<BR>  implicit none<BR>  real(8)  x(3,1),f(3,1)<BR>  f(1,1)=x(1,1)*x(2,1)-x(3,1)**2-1<BR>  f(2,1)=x(1,1)*x(2,1)*x(3,1)+x(2,1)*x(2,1)-x(1,1)**2-2<BR>  f(3,1)=exp(x(1,1))+x(3,1)-exp(x(2,1))-3<BR>endsubroutine