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