Please help me parse the multiplication operation of Pascal s high precision calculations. Thanks 40

Read in by multiplier s1, multiplier s2

Divide S1 and S2 into 4-bit segments, and convert them into numerical values in arrays A and B; Write down the length of a, b, k1, k2;

i is assigned the lowest digit in b;

Take the ith bit from b and multiply it by a, and add it to another array c;

i:=i-1；Detection I value: less than k2 then turn, otherwise turn

Can you not copy the other answers? I want high precision by high precision, big brother.

var i,j,la,lb,len:integer;

s1,s2:string;

m:longint;

a,b,c:array[1..250] of integer;

beginreadln(s1); la:=length(s1);

for i:=1 to la do a[i]:=ord(s1[la-i+1])-48;

readln(s2); lb:=length(s2);

for i:=1 to lb do b[i]:=ord(s2[lb-i+1])-48;

for i:=1 to la do

for j:=1 to lb do

c[i+j-1]:=c[i+j-1]+a[i]*b[j];

len:=la+lb;

for i:=1 to len do

beginc[i+1]:=c[i+1]+c[i] div 10;

c[i]:=c[i] mod 10;

end;while c[len]=0 do dec(len);

m:=c[len];

while m>0 do begin c[len]:=m mod 10;m:=m div 10;inc(len);end;

for i:=len-1 downto 1 do write(c[i]);

end.It should be easy to understand, to put it bluntly, it's just analog multiplication.

Take a look at the following snippets:

Input a, b array, a has n bits, b has m bits, and outputs c array.

fillchar(c,sizeof(c),0) is cleared.

for i:=1 to n do

for j:=1 to m do

c[i+j-1]:=c[i+j-1]+a[i]*b[j];

This sentence is the key, why:

For example, 123*456

Considering the multiplication of the ith digit of the number A and the jth digit of the number b, you might as well take 2 for i and 3 for j, then a[i]=2 a[j]=4 (taken in reverse order) is 20*400

Do you see it, it's a 4-digit number, but it's not necessarily, it could be a 5-digit number, but the common denominator is that there are 3 zeros at the end.

That is. a[i]*10^(i-1)*b[j]*10^(j-1)--a[i]*b[j])*10^(i+j-2)

So it's stored in c[i+j-1].

for i:=1 to n+m do multiply two numbers by the maximum n+m digits.

beginc[i+1]:=c[i+1]+c[i] div 10;

c[i]:=c[i] mod 10;

end;The previous step did not take into account the carrying, and here one wave has processed all the carrying}