P2142 高精度減法
題目描述
高精度減法

分析
高精度減法

var
 s1,s2,s:ansistring;
 a,b,c:array[0..100000]of longint;
 i:longint;
begin
  readln(s1);
  readln(s2);
  a[0]:=length(s1);
  b[0]:=length(s2);
  if (a[0]<b[0])or((a[0]=b[0])and(s1<s2)) then
   begin
     s:=s1;s1:=s2;s2:=s;write('-');
   end;
  a[0]:=length(s1);
  b[0
 
]:=length(s2);
  for i:=a[0] downto 1 do val(s1[i],a[a[0]-i+1]);
  for i:=b[0] downto 1 do val(s2[i],b[b[0]-i+1]);
  for i:=1 to a[0] do
   begin
     c[i]:=a[i]-b[i]+c[i];
     if c[i]<0 then
      begin
        dec(c[i+1]);
        c[i]:=c[i]+10;
      end;
   end;
  for i:=length(s1) downto 2 do if c[i]=0
 
 then dec(a[0]) else break;
  for i:=a[0] downto 1 do write(c[i]);
end.