金额问题：
Function XiaoxieToDaxie(f : String) : String;
var
Fs,dx,d2,zs,xs,h,jg:string;
i,ws,{l,}w,j,lx:integer;
begin
f := Trim(f);
if copy(f,1,1)='-' then begin
Delete(f,1,1);fs:='负';end
else fs:='';
dx:='零壹贰叁肆伍陆柒捌玖';
d2:='拾佰仟万亿';
i := AnsiPos('.',f); //小数点位置
if i = 0 Then
zs := f //整数
else begin
zs:=copy(f,1,i - 1); //整数部分
xs:=copy(f,i + 1,200);
end;
ws:= 0; //l := 0;
for i := Length(zs) downto 1 do begin
ws := ws + 1; h := '';
w:=strtoint(copy(zs, i, 1));
if (w=0) and (i=1) then jg:='零';
If w > 0 Then
Case ws of
2..5:h:=copy(d2,(ws-1)*2-1,2);
6..8:begin
h:=copy(d2,(ws-5)*2-1,2);
If AnsiPos('万',jg)=0 Then h:=h+'万';
end;
10..13:h := copy(d2,(ws-9)*2-1, 2);
End;
jg:=copy(dx,(w+1)*2-1,2) + h + jg;
If ws=9 Then jg := copy(jg,1,2) + '亿' + copy(jg,3,200);
end;
j:=AnsiPos('零零',jg);
While j > 0 do begin
jg := copy(jg, 1, j - 1) + copy(jg, j + 2,200);
j := AnsiPos('零零',jg);
end;
If (Length(jg) > 1) And (copy(jg,length(jg)-1,2)='零') Then jg :=copy(jg,1,Length(jg)-2);
j := AnsiPos('零亿',jg);
If j > 0 Then jg := copy(jg,1, j - 1) + copy(jg, j + 2,200);
//转换小数部分
lx := Length(xs);
If lx > 0 Then begin
jg := jg + '元';
For i := 1 To lx do begin
if i=1 then begin
jg := jg + copy(dx, strtoint(copy(xs,i,1))*2 + 1, 2);
jg := jg +'角';
end;
if i=2 then begin
jg := jg + copy(dx, strtoint(copy(xs,i,1))*2 + 1, 2);
jg := jg +'分';
end;
end;
j :=AnsiPos('零角零分',jg);
if j>0 then jg := copy(jg,1,j-1)+copy(jg,j+8,200)+'整';
j := AnsiPos('零角',jg);
if j>0 then jg := copy(jg,1,j-1)+copy(jg,j+4,200);
j := AnsiPos('零分',jg);
if j>0 then jg := copy(jg,1,j-1)+copy(jg,j+4,200);
End
else
jg := jg + '元整';
result := fs+jg;
end;
