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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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