Pascal high precision multiplication is used to correct mistakes, and good corrections will naturall

This is taught by the teacher in the classroom, and it is highly accurate.

type data=array [0..10000] of longint;

vara,b,c:data;

s,s1:ansistring;

i,l:longint;

procedure ds;

beginreadln(s1);

l:=length(s1);

repeat

s:=copy(s1,l-3,4);

a[0]:=a[0]+1;

val(s,a[a[0]])

delete(s1,l-3,4);

l:=l-4;

until l<=0;

readln(s1);

l:=length(s1);

repeat

s:=copy(s1,l-3,4);

b[0]:=b[0]+1;

val(s,b[b[0]])

delete(s1,l-3,4);

l:=l-4;

until l<=0;

end;procedure cf;

vari,j,len,x:longint;

beginfor i:=1 to a[0] dobeginx:=0;

for j:=1 to b[0] do

beginx:=a[i]*b[j]+x div 10000+c[i+j-1];

c[i+j-1]:=x mod 10000;

end;c[i+j]:=x div 10000;

end;len:=a[0]+b[0];

while (c[len]=0) and (len>1) do dec(len);

c[0]:=len

end;procedure print;

begin;

for i:=c[0] downto 1 dobeginif i<>c[0] then

beginif c[i]<1000 then write(0);

if c[i]<100 then write(0);

if c[i]<10 then write(0);

end;write(c[i]);

end;end;

beginds;

cf;print;

program p1;

vara:array[0..10000] of integer;

b:array[0..10000] of integer;

c:array[0..10000] of integer;

len,x,i,j:integer;

beginreadln(len);

a[0]:=len;

b[0]:=len;

c[0]:=len;

for i:=len downto 1 do read(a[i]);

readln;

for i:=len downto 1 do read(b[i]);

readln;

for i:=1 to len do

for j:=1 to len do

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

c[i+j]:=c[i+j]+c[i+j-1] div 10;

c[i+j-1]:=c[i+j-1] mod 10 ;

end;len:=len*2;

while c[len]=0 do

dec(len);

for i:=len downto 1 dowrite(c[i]);

writeln;

end.— Know that you are welcome to join the team pas world!

Some of the above mistakes may be awarded 100 points without being corrected. This is because the test data may not detect such errors.

Strictly speaking, if there is an example that I gave. Then there is no output. Because there will be no illegal access to A[501].

So it will cause a running error.