Assembly language to add two 100 digit-long numbers and count # of the ones in each one of the two numbers

Hi. I am a beginner of assembly language.

I am so confused on this problem.
I need to write an assembly language program (inline assembler) to add 2, 100 digit-long number each. The same program should be able to count the number of the ones in each one of the two numbers and also in the answer.

Use the numbers:
9876543210123456788908906753421132465798012345678900987654321734561980212345678909087654321908765432
0987654321123456789009876543212314567890123456789012345678900987654321123456789009876543211234567890
to test your program


so the number is 100 digits decimal, that means 32-bit register cannot contain in one time.
I need to divide into array, right?

the program will take this form;

#include<iostream>
using namespace std;

void main()
{
int a[..] = .......................
int b[..] = ......................
int ans;

_asm{


    {

cout << "answer =" << ans;
}

here is the example i got for reference.
To count ones:
int a=5978634, count=0, loop=32;
_asm{
mov eax,a
mov ebx,1
mov ecx,eax
L:
and eax, ebx
shl ebx, 1
cmp eax, 0
je L1
inc count
L1:
mov eax, ecx
dec loop
cmp loop, 0
jg L
}

To subtract 8digit numbers:
int a[2]={1234, 5678}, b[2]={8912,7653}, c, d;
_asm{
mov ebx, 0
mov eax, [a+4]
sub eax, [b+4]
cmp eax, 0
jge L1
neg eax
mov d, eax
mov ebx, 1
L1:
add ebx, b
mov eax, a
sub eax, ebx
mov c, eax
}
cout << c << d << endl;

what i dont understand are that how to put together those, and how to carry the number if there is carry.

Thanks