RT，麻烦大神帮忙编一下

program exam2;
const max=200;
var a,b,c:array[1..max] of 0..9;
n,n1,n2:string;
lena,lenb,lenc,i,x:integer;
begin
write('Input minuend:'); readln(n1); //输入被减数
write('Input subtrahend:');readln(n2); //输入减数
if (length(n1)<length(n2)) or (length(n1)=length(n2)) and (n1<n2) then
begin //处理被减数和减数
n:=n1;
n1:=n2;
n2:=n; 　　 //交换减数和被减数
write('-') //交换了减数和被减数，结果为负数
end;
lena:=length(n1);
lenb:=length(n2);
for i:=1 to lena do
a[lena-i+1]:=ord(n1[i])-ord('0');
for i:=1 to lenb do
b[lenb-i+1]:=ord(n2[i])-ord('0');
i:=1;
x:=0;
while (i<=lena) or(i<=lenb) do
begin
if a[i]<b[i] then //不够减，那么向高位借1当10
begin
a[i]:=a[i]+10;
a[i+1]:=a[i+1]-1;
end;
c[i] := a[i]-b[i] ; //对应位相减
i := i + 1
end;
lenc:=i;
while (c[lenc]=0) and (lenc>1) do dec(lenc); //最高位的0不输出
for i:=lenc downto 1 do
write(c[i]);
end.
减法

const max=200;
var a,b,c:array[1..max] of 0..9;
n1,n2:string;
lena,lenb,lenc,i,j,x:integer;
begin
write('Input multiplier:'); readln(n1); write('Input multiplicand:'); readln(n2);
lena:=length(n1); lenb:=length(n2);
for i:=1 to lena do a[lena-i+1]:=ord(n1[i])-ord('0');
for i:=1 to lenb do b[lenb-i+1]:=ord(n2[i])-ord('0');
for i:=1 to lena do
begin
x:=0; //用于存放进位
for j:=1 to lenb do
begin //对乘数的每一位进行处理
c[i+j-1] := a[i]*b[j] + x + c[i+j-1]; //当前乘积+上次乘积进位+原数
x:=c[i+j-1] div 10; c[i+j-1] := c[i+j-1] mod 10;
end;
c[i+j]:= x; 　　　 //进位
end;
lenc:=i+j;
while (c[lenc]=0) and (lenc>1) do dec(lenc);
for i:=lenc downto 1 do write(c[i]);
writeln
end.
乘法

因懒得写从课件Ctrl v过来的