fortran90源程序最小二乘法解多项式拟合问题.doc

最小二乘法解多项式拟合问题 doubleprecision a(100,100),c(100,100),d(100),alpha(100) doubleprecision gs(100,100),temp,q,delt(100),x(100),y(100) integergsn open(100,file=in.txt,status=old) read(100,*)n,m read(100,*)(x(i),i=1,n),(y(j),j=1,n) close(100) do1i=1,n do2j=1,m+1 910 a(i,j)=x(i)*(j-1) 2 continue 1 continue open(200,file=process.txt,status=unknown) write(200,*)a(m,n):,m:,n,n:,(m+1), do3i=1,n write(200,35)(a(i,j),j=1,(m+1) 3 continue do4i=1,m+1 do7j=1,m+1 920 c(i,j)=0 do8k=1,n c(i,j)=c(i,j)+x(k)*(i+j-2) 8 continue 7 continue 4 continue do9i=1,m+1 d(i)=0 do11k=1,n 929 d(i)=d(i)+y(k)*(x(k)*(i-1) 11 continue 9 continue write(200,*)c(n,n):,n:,(m+1), do12i=1,m+1 write(200,35)(c(i,j),j=1,m+1) 12 continue write(200,*)d(n):,n:, write(200,35)(d(i),i=1,m+1) close(200) loop=0 940 gsn=m+1 open(300,file=equation.txt,status=unknown) write(300,*)Theaddedmatrixis: do13i=1,gsn write(300,35)(c(i,j),j=1,gsn),d(i) 13 continue do14i=1,gsn do15j=1,gsn gs(i,j)=c(i,j) 15 continue 950 gs(i,gsn+1)=d(i) 14 continue do16k=1,gsn+1 do17i=k+1,gsn+1 if(dabs(gs(k,k).lt.dabs(gs(i,j)then do18j=1,gsn+1 temp=gs(k,j) gs(k,j)=gs(i,j) gs(i,j)=temp 18continue endif 17continue temp=gs(k,k) do19j=K,gsn+1 gs(k,j)=gs(k,j)/temp 19 continue do21i=k,gsn+1 temp=gs(i,j) do22j=k,gsn+1 gs(i,j)=gs(i,j)-temp*gs(k,j) 22 continue 21 continue write(300,*)- loop=loop+1 write(300,*)loop:,loop do23i=1,m write(300,35)(gs(i,j),j=1,gsn+1) 23 continue 16 continue alpha(gsn)=gs(gsn,gsn+1) do24k=gsn-1,1,-1 temp=0.0d0 do25j=k+1,gsn temp=temp+gs(k,j)*alpha(j) 25 continue 24 continue write(300,*)- write(300,35)(alpha(i),i=1,gsn) q=0.0d0 do26i=1,n delt(i)=-y(i) do27j=1,m+1 delt(i)=delt(i)+(x(i)*(j-1)*alpha(j) 27 continue q=q+delt(i)*2 26 continue write(300,*)- write(300,*)Thefinalresult: write(300,31)(x(i),i=1,n) write(300,32)(y(i),i=1,n) write(300,33)(delt(i),i=1,n) write(300,*)q:,q close(300) write(*,*)(alpha(i),i=1,m+1) 31 format