回复：我又要开始奋斗了。

var s:string; n,i,j,k:longint;
     a:array[0..100]of string; t1,t2:string;
procedure insert(x:string);
begin
inc(k); a[k]:=x;
end;
function pop:string;
begin
pop:=a[k];
dec(k);
end;
procedure init;
begin
readln(s); n:=length(s);
k:=0;
for i:=1 to 100 do a[i]:='';
for i:=1 to n do
   begin
    if not(s[i] in ['+','-','*','/']) then insert(s[i])
    else
     begin
      t1:=pop; t2:=pop;
      if(length(t1)=1)and(length(t2)=1) then insert(t2+s[i]+t1)
      else if(length(t1)>1)and(length(t2)=1) then
       insert(t2+s[i]+'('+t1+')')
       else if(length(t1)=1)and(length(t2)>1) then
              insert('('+t2+')'+s[i]+t1)
        else if(length(t1)>1)and(length(t2)>1)   then
              insert('('+t2+')'+s[i]+'('+t1+')');
     end;
   end;
end;
procedure print;
begin
for i:=1 to k do write(a[i]);
end;
begin
init;
print;
end.
-----------
后缀表达式转换成中缀。。。我没设优先级。。。成功的把一道堆栈的题改成了二分去括号………………

var s1,s2,t:string;
function bj(a,b:string):boolean;
var i:integer;
begin
   if length(a)>length(b) then exit(true)
   else
    if length(a)<length(b) then exit(false)
    else
     for i:=1 to length(a) do if a[i]>b[i] then exit(true)
     else if a[i]<b[i] then exit(false);
    exit(true);
end;
procedure js(s1,s2:string);
var a,b:array[0..201]of integer;
     j,w,i:longint;
begin
   fillchar(a,sizeof(a),0);
   fillchar(b,sizeof(b),0);
   w:=length(s1);
   for i:=1 to w do a[w-i+1]:=ord(s1[i])-48;
   for i:=1 to length(s2) do b[length(s2)-i+1]:=ord(s2[i])-48;
   for i:=1 to w do
    begin
     a[i]:=a[i]-b[i];
     if a[i]<0 then begin inc(a[i],10); dec(a[i+1]); end;
    end;
    j:=1;
//    while a[j]=0 do inc(j);
while a[w]=0 do dec(w);
for i:=w downto 1 do write(a[i]);
end;
begin
   readln(s1);
   readln(s2);
    if not bj(s1,s2) then
    begin write('-'); t:=s1; s1:=s2; s2:=t; end;
    js(s1,s2);
end.
-------------
高精度减，，不知道怎么蛋疼的写了这么长

var s1,s:string; i,j,n:longint;
     a:array[0..1000]of integer;
     c:char;b:array['a'..'z']of integer;
procedure init;
begin
read(c);
while c<>'.' do
   begin
    s:=s+c;
    read(c);
   end;
s:=s+c;
n:=length(s);
s1:=lowercase(s);
for i:=1 to n do write(s1[i]);
writeln;
end;
procedure main;
begin
j:=0;
for i:=1 to n do
   if s[i]=' ' then   inc(j)
   else begin a[i-1]:=j; j:=0; end;
   for i:=1 to n do
    if(a[i]=0)and(s[i]<>' ') then write(s[i])
    else if a[i]>0 then write(' ');
    j:=1;
   WRITELN;
   for i:=2 to n do
    begin
    if( s[i] in ['A'..'Z'])or(s[i] in ['a'..'z']) then
     if lowercase(s[i])=lowercase(s[i-1]) then inc(j)
     else begin write(lowercase(s[i-1]));if j<>1 then write(j); j:=1; end;
     if i=n then begin write(lowercase(s[i-1]));if j<>1 then write(j); end;
    end;
end;
begin
init;
main;
end.
1个字符串处理

户口本？
他不是在和腾讯合作一个项目么………………
你和他同学？？？
当年老罗就拿他勾引我来学信息的。。在机房膜拜过胡神牛。。