Qfyilyi

I'm just a beginner in Assembly MIPS. I have tried to write a program in MIPS to divide two integers below:

#Data segment

    .data

Input1: .asciiz "Dividend (16 bit): "       

Input2: .asciiz "Divisor (8 bit): "     

OutQ:   .asciiz "Quotient (8 bit) = "       

OutR:   .asciiz "Remainder (8 bit) = "      

#Code segment

    .text

    .globl  main

main:   



    la  $a0,Input1

    addi    $v0,$zero,4

    syscall

    addi    $v0,$zero,5

    syscall

    addu    $t2,$zero,$v0

    la  $a0,Input2

    addi    $v0,$zero,4

    syscall

    addi    $v0,$zero,5

    syscall

    addu    $t3,$zero,$v0

#Initiate Q=0, R=divisor, D<<7, Cnt=8

    sll $t3,$t3,7       #create Divisor*(2^7)

    addi    $t0,$zero,0 #Q=0

    addu    $t1,$zero,$t2   #R=divisor

    addi    $t4,$zero,8 #R = R - D

divide0:    sub $t1,$t1,$t3 

#R<0?

    lui $t5,0x8000

    and $t5,$t5,$t1

    beq $t5,$zero,Divide1

# R<0

    add $t1,$t1,$t3 

    sll $t0,$t0,1       #bit_i=0

    b   Divide2

# R>=0

Divide1:    sll $t0,$t0,1       

    ori $t0,$t0,1       #bit_i=1

#Shift Right D

Divide2:    srl $t3,$t3,1       

#n=8 times  

    addi    $t4,$t4,-1

    bne $t4,$zero,divide0

#Print Result

    la  $a0,OutQ

    addi    $v0,$zero,4

    syscall

    addu    $a0,$zero,$t0

    addi    $v0,$zero,1

    syscall

    addu    $a0,$zero,'n'

    addi    $v0,$zero,11

    syscall

    la  $a0,OutR

    addi    $v0,$zero,4

    syscall

    addu    $a0,$zero,$t1

    addi    $v0,$zero,1

    syscall

#End Program

    ADDI    $v0,$zero,10

    syscall

I wonder whether I can modify this program to divide two floating point numbers without using div.s and how can I do it?
I am just a beginner so please help me with this!
Thank you so much in advance!