SUBROUTINE SGTSL(N,C,D,E,B,INFO)
      INTEGER N,INFO
      REAL C(1),D(1),E(1),B(1)
C
C     SGTSL GIVEN A GENERAL TRIDIAGONAL MATRIX AND A RIGHT HAND
C     SIDE WILL FIND THE SOLUTION.
C
C     ON ENTRY
C
C        N       INTEGER
C                IS THE ORDER OF THE TRIDIAGONAL MATRIX.
C
C        C       REAL(N)
C                IS THE SUBDIAGONAL OF THE TRIDIAGONAL MATRIX.
C                C(2) THROUGH C(N) SHOULD CONTAIN THE SUBDIAGONAL.
C                ON OUTPUT C IS DESTROYED.
C
C        D       REAL(N)
C                IS THE DIAGONAL OF THE TRIDIAGONAL MATRIX.
C                ON OUTPUT D IS DESTROYED.
C
C        E       REAL(N)
C                IS THE SUPERDIAGONAL OF THE TRIDIAGONAL MATRIX.
C                E(1) THROUGH E(N-1) SHOULD CONTAIN THE SUPERDIAGONAL.
C                ON OUTPUT E IS DESTROYED.
C
C        B       REAL(N)
C                IS THE RIGHT HAND SIDE VECTOR.
C
C     ON RETURN
C
C        B       IS THE SOLUTION VECTOR.
C
C        INFO    INTEGER
C                = 0 NORMAL VALUE.
C                = K IF THE K-TH ELEMENT OF THE DIAGONAL BECOMES
C                    EXACTLY ZERO.  THE SUBROUTINE RETURNS WHEN
C                    THIS IS DETECTED.
C
C     LINPACK. THIS VERSION DATED 08/14/78 .
C     JACK DONGARRA, ARGONNE NATIONAL LABORATORY.
C
C     NO EXTERNALS
C     FORTRAN ABS
C
C     INTERNAL VARIABLES
C
      INTEGER K,KB,KP1,NM1,NM2
      REAL T
C     BEGIN BLOCK PERMITTING ...EXITS TO 100
C
         INFO = 0
         C(1) = D(1)
         NM1 = N - 1
         IF (NM1 .LT. 1) GO TO 40
            D(1) = E(1)
            E(1) = 0.0E0
            E(N) = 0.0E0
C
            DO 30 K = 1, NM1
               KP1 = K + 1
C
C              FIND THE LARGEST OF THE TWO ROWS
C
               IF (ABS(C(KP1)) .LT. ABS(C(K))) GO TO 10
C
C                 INTERCHANGE ROW
C
                  T = C(KP1)
                  C(KP1) = C(K)
                  C(K) = T
                  T = D(KP1)
                  D(KP1) = D(K)
                  D(K) = T
                  T = E(KP1)
                  E(KP1) = E(K)
                  E(K) = T
                  T = B(KP1)
                  B(KP1) = B(K)
                  B(K) = T
   10          CONTINUE
C
C              ZERO ELEMENTS
C
               IF (C(K) .NE. 0.0E0) GO TO 20
                  INFO = K
C     ............EXIT
                  GO TO 100
   20          CONTINUE
               T = -C(KP1)/C(K)
               C(KP1) = D(KP1) + T*D(K)
               D(KP1) = E(KP1) + T*E(K)
               E(KP1) = 0.0E0
               B(KP1) = B(KP1) + T*B(K)
   30       CONTINUE
   40    CONTINUE
         IF (C(N) .NE. 0.0E0) GO TO 50
            INFO = N
         GO TO 90
   50    CONTINUE
C
C           BACK SOLVE
C
            NM2 = N - 2
            B(N) = B(N)/C(N)
            IF (N .EQ. 1) GO TO 80
               B(NM1) = (B(NM1) - D(NM1)*B(N))/C(NM1)
               IF (NM2 .LT. 1) GO TO 70
               DO 60 KB = 1, NM2
                  K = NM2 - KB + 1
                  B(K) = (B(K) - D(K)*B(K+1) - E(K)*B(K+2))/C(K)
   60          CONTINUE
   70          CONTINUE
   80       CONTINUE
   90    CONTINUE
  100 CONTINUE
C
      RETURN
      END