sub fwdsub.for

来自「非线性回归问题SQP解法」· FOR 代码 · 共 19 行

	SUBROUTINE FWDSUB(N,NAMAX,D,TRANS,W)
	DOUBLE PRECISION D(NAMAX,N),W(N)
	LOGICAL TRANS
	IF (TRANS)GOTO 3
	DO 2 I=1,N
		DO 1 J=1,I-1
			W(I)=W(I)-D(I,J)*W(J)
1		CONTINUE
		W(I)=W(I)/D(I,I)
2	CONTINUE
	RETURN
3	DO 5 I=1,N
		DO 4 J=1,I-1
			W(I)=W(I)-D(I,J)*W(J)
4		CONTINUE
		W(I)=W(I)/D(I,I)
5	CONTINUE
	RETURN
	END