Re: May be, I am too.
var
c,counter,n,i:integer;
a:array[1..100] of integer;
begin
readln(n); c:=9;
if n=0 then write('10');
if n=1 then write('0');
repeat
if (n mod c=0) then
begin
n:=n div c;
inc(counter);
a[counter]:=c;
end
else c:=c-1;
until c=1;
if n<>1 then write('-1');
for i:=counter downto 1 do
write(a[i]);
end.
I think that answer is true
can you say why is this false
haa-----
can you translate 1002 and 1003 problems to me in Azeri
or in turkish languages
or explain form in english
I can't understand much more in english
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
I think, that test #4 incorrect. My program is simple.
It take the lager divizor of N, that less then 10, and put
it at begin of Q. If after that N != 1, then write -1. It's easy to prove, that it's true solution, but it take WA.
