! EXAM 44.
!   Hydrogen fluoride hexamer...(HF)6
!   using the divide-and-conquer (DC) method
!
!  Divide-and-conquer HF and MP2 energies are:
!     FINAL DC-RHF E= -599.9687803934, 12 iters
!             E(MP2)= -600.7532099625
!            SCS-MP2= -600.7342987431
!  Selecting only DC-MP2, with full SCF accuracy:
!              RHF E= -599.9690828876, 10 iters
!              MP2 E= -600.7535223025
!            SCS-MP2= -600.7346097076
!  An explicit MP2/6-31G calculation yields:
!              MP2 E= -600.7532239374
!            SCS-MP2= -600.7343475232
!
!  CCSD calculation requires changing one keyword in
!  $CONTRL.  The divide-and-conquer CCSD energy is
!        CCSD ENERGY= -600.7629001759
!  Selecting only DC-CCSD, with full SCF accuracy:
!        CCSD ENERGY= -600.7640339587
!  compared to the explicit CCSD/6-31G calculation:
!        CCSD ENERGY= -600.7632011798
!
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY MPLEVL=2 COORD=ZMT $END
 $SYSTEM MWORDS=1 $END
 $BASIS  GBASIS=N31 NGAUSS=6 $END
 $GUESS  GUESS=HUCSUB $END
 $SCF    DIRSCF=.TRUE. DIIS=.TRUE. $END
!   the subsystems consist of individual HF monomers
 $DANDC  DCFLG=.TRUE. BUFRAD=5.0 BUFTYP=RADSUB
         NSUBS=6 LBSUBS(1)=1,2,3,4,5,6,1,2,3,4,5,6 $END
 $DCCORR DODCCR=.TRUE. RBUFCR=3.0 $END
 $DATA
zigzag hexamer (HF)6
C1
F
F  1 rFF
F  2 rFF  1 aHFH
F  3 rFF  2 aHFH  1 180.0
F  4 rFF  3 aHFH  2 180.0
F  5 rFF  4 aHFH  3 180.0
H  1 rHFL 2 aHFH  3 180.0
H  2 rHF  3 aHFH  4 180.0
H  3 rHF  4 aHFH  5 180.0
H  4 rHF  5 aHFH  6 180.0
H  4 rFH  3 aHFH  2 180.0
H  5 rFH  4 aHFH  3 180.0

rFF=2.5
rHF=0.97
rFH=1.53
aHFH=116.0
rHFL=0.97
 $END

Link to the log file of this example.
exam44 Log File