Question

求斐波那契(Fibonacci)数列的第 10 项,已知该数列的前两项都为 1,即 F(1)=1,F(2)=

Answer 1

斐波那契(Fibonacci)数列的通项公式为：
an=(an-1)+a(n-2),(n≥3 n∈N*)
∴斐波那契数列的前十项为：
1 1 2 3 5 8 13 21 34 55
第十项为55

Answer 2

stack segment stack 'stack'
dw 32 dup(0)
stack ends
data segment
A DB 8 DUP(0)
OBF DB 3,0,3 DUP(0)
data ends
code segment
begin proc far
assume ss:stack,cs:code,ds:data
push ds
sub ax,ax
push ax
mov ax,data
mov ds,ax

MOV DX,OFFSET OBF
MOV AH,10
INT 21H
MOV AL,OBF+1
CMP AL,1
JA NEX2
MOV CL,OBF+2
AND CL,0FH
JMP NEX1
NEX2: MOV AL,OBF+2
AND AL,0FH
MOV CL,10
MUL CL
MOV AH,OBF+3
AND AH,0FH
ADD AL,AH
MOV CL,AL

NEX1: MOV CH,0
CMP CX,3
JB ONE
SUB CX,2

MOV AX,1
MOV BX,1
AGAIN: MOV DX,AX
ADD AX,BX
MOV BX,DX
LOOP AGAIN

MOV BX,7
MOV A[BX],'$'
MOV CX,10
AG: MOV DX,0
DIV CX
OR DL,30H
DEC BX
MOV A[BX],DL
AND AX,AX
JNZ AG
SUB BX,2
MOV WORD PTR A[BX],0A0DH
MOV DX,BX
ADD DX,OFFSET A
MOV AH,9
INT 21H
JMP LAST

ONE: MOV DL,0AH
MOV AH,2
INT 21H
MOV DL,31H
MOV AH,2
INT 21H

LAST: RET
begin endp
code ends
end begin

Answer 3

55

Answer 4

1 1 2 3 5 8 13 21 34 55