 
m1 Вычитание
чисел с плавающей
точкой
    
RDYOUT:=1; ERROR:=1; PR:=1
Логический
сдиг слова
вправо на 1 или
2 разряда
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d1
Курсовую
роботу сдавал
24.05.95 2:5020/403.34 (2:5020/235.27)
RDYIN==1 да
Преподаватель:
Иваненко :-(
нет
  
  
( A1[0...7] ):=( II[4...11] ); B1:=II[0..3] )
m2 A1[0...7]:=( A1[1..7],B1[0] ) m22
B1[0..3]:=(B1[1..3],A1[0])
C=C+1
m9
d2
d15
 
A1[1]==1 нет
A1[3]==1
да
да
нет
( k1,k2) = mod3(A1[0..7]) + mod3(C )
(
A2[0...7] ):=( II[4...11] ); B2:=II[0...3] ) m3
A1[0...7]:=( A1[1..7],B1[0] ) m23
 
B1[0..3]:=(B1[1..3],A1[0])
  (
P1,S1[0...4] ) = ( B1[0],B1[0...3] ) - ( B2[0],B2[0...3] );C=A2
m4
(P2,S2[0...8] )=(A1[0],A1[0...7] ) - (A2[0],A2[0...7] ) m10
 
d3 да
( k3,k4) =mod3( p2,S2[0...8] )
  
S1[0] + S1[1]
нет
d9
 
нет
(k1,k2)==(k3,k4)
d4 да
1 0
да
S1[1...4] ==0
S1[0]
d10
d5
S2[0] + S2[1]
нет
да
  
0 S1[0] 1
d6
(
A1[0...7] ):=( S2[1...8] ) m13
p1,S1[0...4]=( B1[0...3] ) +1
 
m11
d12
d11
  
B2:=B2+1 m5 B1:=B1+1
m6 да
A1==0 нет
S1[0] + S1[1]
A2[0...7]:=(
A2[0],A2[0...6] ) A1[0...7]:=( A1[0],A1[0...6] )
d13
да
 
A1[1] + A1[0] нет
да
  
A2==0 нет
A1==0
B1:=0 m14
нет (A1[0...7] ):=(
S2[0...7] )
d7 да
d8
m15 p1,S1[0...4]=(
B1[0],B1[0...3] ) - 1 B1:=S1[1...4]
да
m12
d14
A1:=0 m7
S1[0] +
S1[1]
A1:=-A2;B1:=B2 m8
да
нет
B1:=0
B1:=S1[1...4]
  
A1:=0 m16
m17 A1[0...7]:=( A1[0..6],0 )
 
RDYOUT:=0 m20
m19
  
ERROR:=0 m18 PR:=0
IO[0...11]
= ( A1[0...7],B1[0...3] ) m21
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