[Fortran] 纯文本查看 复制代码
PROGRAM MAIN !主程序
IMPLICIT NONE
INTEGER,PARAMETER::POINTS_NUM=15 !这个是测试点的数目，你应该根据你的测试点数修改它
INTEGER,PARAMETER::POLY_ORDER=13 !如果是线性拟合：POLY_ORDER=1
INTEGER,PARAMETER::DP=KIND(1.D0)
REAL(KIND=DP)::A(POINTS_NUM,POLY_ORDER+1),POLY_COEF(POLY_ORDER+1) !如果是线性拟合，实际上就是POLY_COEF(2)：y = ax + b
REAL(KIND=DP)::XT(POINTS_NUM),YT(POINTS_NUM),DX,XN(POINTS_NUM),YN(POINTS_NUM)
INTEGER::INFO
INTEGER::I
!XT,YT为测试的数据
XT=(/0.1D0,0.2D0,0.3D0,0.4D0,0.5D0,0.6D0,0.7D0,0.8D0,0.9D0,0.92D0,0.94D0,0.96D0,0.98D0,0.99D0,0.995D0/)!插值节点
YT=(/0.86D0,0.78D0,0.71D0,0.66D0,0.59D0,0.52D0,0.45D0,0.36D0,0.23D0,0.19D0,0.16D0,0.12D0,0.07D0,0.04D0,0.02D0/)!插值节点函数值
!XN=(/0.15D0,0.985D0,0.995D0/)!待计算的插值点
!YN=(/0.86060402972635042,5.30099995722376274E-002,2.57654880479794013E-002/)!插值结果
DX=1.D0/15
CALL LSINTERPOLY(POINTS_NUM,XT,YT,POLY_ORDER,POLY_COEF,INFO)
WRITE(*,*)
WRITE(*,*)
IF(INFO/=-1)THEN
DO I=1,15
XN(I)=XT(I)
CALL POLY_VALUE(POLY_ORDER,POLY_COEF,XN(I),YN(I))
!WRITE(*,FMT="(4X,A,G0,A,G0)")"YN(",XN(I),")=",YN(I)
WRITE(*,FMT="(F6.4,2X,G0)")YN(I),YN(I)-YT(I)
END DO
ELSE
WRITE(*,*)"ERROR"
ENDIF
READ(*,*)
END PROGRAM MAIN
!---------------------------------------------------------------------------
SUBROUTINE LSINTERPOLY(POINTS_NUM& !测试数据的个数
,XT& !测试数据横坐标
,YT& !测试数据纵坐标
,POLY_ORDER& !多项式阶数
,POLY_COEF& !多项式系数
,INFO& !输出信息，INF=0，求解成功，INFO=-1求解失败
)!求多项式系数
IMPLICIT NONE
INTEGER,PARAMETER::DP=KIND(1.D0)
INTEGER,INTENT(IN)::POINTS_NUM,POLY_ORDER
REAL(KIND=DP),INTENT(IN)::XT(POINTS_NUM),YT(POINTS_NUM)
REAL(KIND=DP),INTENT(OUT)::POLY_COEF(POLY_ORDER+1)
INTEGER,INTENT(OUT)::INFO
LOGICAL::SOL_OK
REAL(KIND=DP)::A(POINTS_NUM,POLY_ORDER+1),ATA(POLY_ORDER+1,POLY_ORDER+1)
REAL(KIND=DP)::B(POINTS_NUM),C(POLY_ORDER+1,POINTS_NUM)
INTEGER::I,J
IF(POLY_ORDER+1>POINTS_NUM)THEN
INFO=-1
RETURN
END IF
INFO=0
A=0.D0
DO I=1,POINTS_NUM
DO J=0,POLY_ORDER
IF(J==0)THEN
A(I,J+1)=1.D0
ELSE
A(I,J+1)=XT(I)**J
ENDIF
END DO
END DO
B=YT
C=TRANSPOSE(A)
ATA=MATMUL(C,A)
POLY_COEF=MATMUL(C,B)
! IF(POLY_ORDER+1==POINTS_NUM)THEN
! CALL SOLVE(A,POLY_ORDER+1,POLY_COEF,SOL_OK)
! ELSE
! CALL SOLVE(ATA,POLY_ORDER+1,POLY_COEF,SOL_OK)
! END IF
CALL SOLVE(ATA,POLY_ORDER+1,POLY_COEF,SOL_OK)
IF(SOL_OK)THEN
INFO=0
ELSE
INFO=-1
END IF
RETURN
END SUBROUTINE
!---------------------------------------------------------------------------
SUBROUTINE POLY_VALUE(POLY_ORDER,POLY_COEF,X,Y)!根据多项式计算插值点(X)的函数值Y
IMPLICIT NONE
INTEGER,PARAMETER::DP=KIND(1.D0)
INTEGER,INTENT(IN)::POLY_ORDER
REAL(KIND=DP),INTENT(IN)::POLY_COEF(POLY_ORDER+1),X
REAL(KIND=DP),INTENT(OUT)::Y
INTEGER::I
Y=0.D0
DO I=0,POLY_ORDER
Y=Y*X+POLY_COEF(POLY_ORDER+1-I)
END DO
RETURN
END SUBROUTINE
!---------------------------------------------------------------------------
SUBROUTINE SOLVE(A,N,B,SOL_OK)!全选主元高斯消元回带法求解器
IMPLICIT NONE
INTEGER,PARAMETER::DP=KIND(1.D0)
INTEGER,INTENT(IN)::N
LOGICAL,INTENT(OUT)::SOL_OK
REAL(KIND=DP),INTENT(INOUT)::A(N,N)
REAL(KIND=DP),INTENT(INOUT)::B(N)
REAL(KIND=DP)::BIG,DUM,PIVINV,TEMP
INTEGER I,ICOL,IROW,J,K,L,LL,INDXC(N),INDXR(N),IPIV(N)
SOL_OK=.TRUE.
DO J=1,N
IPIV(J)=0
ENDDO
DO I=1,N
BIG=0.D0
DO J=1,N
IF(IPIV(J).NE.1)THEN
DO K=1,N
IF (IPIV(K).EQ.0) THEN
IF (ABS(A(J,K)).GE.BIG)THEN
BIG=ABS(A(J,K))
IROW=J
ICOL=K
ENDIF
ELSEIF (IPIV(K).GT.1) THEN
WRITE(*,*)'SINGULAR MATRIX IN GAUSSJ'
SOL_OK=.FALSE.
ENDIF
ENDDO
ENDIF
ENDDO
IPIV(ICOL)=IPIV(ICOL)+1
IF (IROW.NE.ICOL) THEN
DO L=1,N
DUM=A(IROW,L)
A(IROW,L)=A(ICOL,L)
A(ICOL,L)=DUM
ENDDO
DUM=B(IROW)
B(IROW)=B(ICOL)
B(ICOL)=DUM
ENDIF
INDXR(I)=IROW
INDXC(I)=ICOL
TEMP=ABS(A(ICOL,ICOL))
IF(A(ICOL,ICOL)==0.D0)THEN
WRITE(*,*) 'SINGULAR MATRIX IN GAUSSJ'
SOL_OK=.FALSE.
RETURN
ENDIF
PIVINV=1.D0/A(ICOL,ICOL)
A(ICOL,ICOL)=1.D0
DO L=1,N
A(ICOL,L)=A(ICOL,L)*PIVINV
ENDDO
B(ICOL)=B(ICOL)*PIVINV
DO LL=1,N
IF(LL.NE.ICOL)THEN
DUM=A(LL,ICOL)
A(LL,ICOL)=0.D0
DO L=1,N
A(LL,L)=A(LL,L)-A(ICOL,L)*DUM
ENDDO
B(LL)=B(LL)-B(ICOL)*DUM
ENDIF
ENDDO
ENDDO
DO L=N,1,-1
IF(INDXR(L).NE.INDXC(L))THEN
DO K=1,N
DUM=A(K,INDXR(L))
A(K,INDXR(L))=A(K,INDXC(L))
A(K,INDXC(L))=DUM
ENDDO
ENDIF
ENDDO
RETURN
END SUBROUTINE SOLVE