[转帖]victory:fortran 语言实现
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<=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))<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<=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
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