Применение неэмпирических и полуэмпирических методов в квантово-химических расчетах. Кобзев Г.И. - 101 стр.

 TWO ELECTRON INTEGRAL EVALUATION REQUIRES 34529 WORDS
OF MEMORY.
II,JST,KST,LST = 1 1 1 1 NREC =   1 INTLOC = 1
II,JST,KST,LST = 2 1 1 1 NREC =   1 INTLOC = 2
II,JST,KST,LST = 3 1 1 1 NREC =   1 INTLOC = 7
II,JST,KST,LST = 4 1 1 1 NREC =   1 INTLOC = 22
II,JST,KST,LST = 5 1 1 1 NREC =   1 INTLOC = 56
II,JST,KST,LST = 6 1 1 1 NREC =   1 INTLOC = 121
II,JST,KST,LST = 7 1 1 1 NREC =   1 INTLOC = 220
II,JST,KST,LST = 8 1 1 1 NREC =   1 INTLOC = 496
II,JST,KST,LST = 9 1 1 1 NREC =   1 INTLOC = 1057
II,JST,KST,LST = 10 1 1 1 NREC =   1 INTLOC = 1577
II,JST,KST,LST = 11 1 1 1 NREC =   1 INTLOC = 2276
II,JST,KST,LST = 12 1 1 1 NREC =   1 INTLOC = 3234
II,JST,KST,LST = 13 1 1 1 NREC =   1 INTLOC = 4420
II,JST,KST,LST = 14 1 1 1 NREC =   2 INTLOC = 871
II,JST,KST,LST = 15 1 1 1 NREC =   2 INTLOC = 4932
II,JST,KST,LST = 16 1 1 1 NREC =   4 INTLOC = 783
TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS =
23801
        5 INTEGRAL RECORDS WERE STORED ON DISK FILE 8.
...... END OF TWO-ELECTRON INTEGRALS .....

CPU    TIME: STEP = 0.24 , TOTAL =     0.4 SECONDS ( 0.0 MIN)
WALL CLOCK TIME: STEP =    0.24 , TOTAL =    0.3 SECONDS ( 0.0
MIN)
CPU UTILIZATION: STEP = 102.06%, TOTAL = 119.55%

    --------------------
    ROHF SCF CALCULATION
    --------------------

  NUCLEAR ENERGY =        28.2227866261
  MAXIT =300 NPUNCH= 2 MULT= 3
  EXTRAP=T DAMP=F SHIFT=F RSTRCT=F DIIS=F SOSCF=T
  DENSITY CONV= 1.00E-05
  ROHF CANONICALIZATION PARAMETERS
     C-C O-O V-V
  ALPHA -0.5000 0.5000 1.5000
  BETA 1.5000 0.5000 -0.5000
  SOSCF WILL OPTIMIZE 185 ORBITAL ROTATION ANGLES.
SOGTOL= 2.500E-01
  MEMORY REQUIRED FOR UHF/ROHF STEP= 15759 WORDS.



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