m1 Вычитание чисел с плавающей точкой
RDYOUT:=1; ERROR:=1; PR:=1 Логический сдиг слова вправо на 1 или 2 разряда
(Учащимся в МИРЭА посвещается!)
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?????
????????????? ?????????????? ?????????????
????????
A1[0-3]
S[0-27]
S[0-3]
S[4-27]
p
A0
XY[0-23]
z12
SRDY
SANSWR
A3
A2
A1
B[0-23]
A2[0-23]
z
B[0-23]
Y[0-23]
Y[8-11]
Y[4-7]
Y[0-3]
B[0-23]
A’[0-23]=A1[3],A2[0-22]
A2[0-23]
B[0-23]
A2[0-23]
A1[0-3]
A1[0-3]
A1[1-3]
X[0-23]
X[0-23]
1
Ix24
&
c
INC
“0”
CW
“1”