check if any char is a digit

My AC program gives that answers for your tests
120: 15
1081: 46
1073845: 1465
500500: 1000
11325: 150
17391: 186
122750946: 15668
906531: 1346
154290: 555
7620753696: 123456

Edited by author 24.01.2018 22:46

Do you know how floating-point numbers work?

10^600 gives "OverflowError: int too large to convert to float" :)

You can use Python library "decimal"

:DD

I found a mistake, it was that: if you use java, you must use BigInteger, not just long or double. I hope it will help someone.

Oh God!
Do you really think that extended holds 600 digits?
You should implement that formula on big numbers...

Thank you VERY much! :-)

Edited by author 10.02.2010 23:16

Use the big basis, for example 1000000000

Java BigInteger BS worked in 0.078ms! Fantastic!

The answer is easier: N = [sqrt(2*M)]

type ac=array[1..601]of longint;
var a,c,mi,lan,ch:ac;
b,d,e,f,g,h,i,j,k,l,m,n:longint;t:array[1..601]of char;
function times(a,c:ac;n:longint):ac;
var i,j,l:longint;
begin
 fillchar(ch,sizeof(ch),0);
 for i:=n downto 1 do begin
  l:=0;
  for j:=n downto 1 do
   if i+j-n>0 then begin
    ch[i+j-n]:=ch[i+j-n]+a[i]*c[j]+l;
    l:=ch[i+j-n] div 10;
    ch[i+j-n]:=ch[i+j-n] mod 10;
   end;
 end;
 times:=ch;
end;
function larger(a,c:ac;n:longint):boolean;
var i:longint;
begin
 for i:=1to n do
  if a[i]>c[i] then exit(true)
  else if a[i]<c[i] then exit(false);
 exit(false);
end;
procedure goal(g:longint);
var i,j:longint;
begin
 if g=602 then begin
  for i:=601-m+1 to 601 do
   write(c[i]);
  WRITELN;
  halt;
 end;
 fillchar(lan,sizeof(lan),0);
 if odd(n) then
  for j:=2to (g-(601-m))*2 do
   LAN[j]:=a[j+(601-n)-1]
 else
  for j:=1to (g-(601-m))*2 do
   LAN[j]:=a[j+(601-n)];
 for i:=1to 10 do begin
  if i<10 then begin
   c[g]:=i;
   fillchar(mi,sizeof(mi),0);
   for j:=1to g-(601-m) do
    mi[j+g-(601-m)]:=c[j+(601-m)];
   mi:=times(mi,mi,(g-(601-m))*2);
  end;
  if i=10 then begin
   c[g]:=9;
   goal(g+1);
  end
  else if larger(mi,lan,(g-(601-m))*2) then begin
   c[g]:=i-1;
   goal(g+1);
  end;
 end;
end;
begin
 n:=0;
 repeat
  inc(n);
  read(t[n]);
 until eoln;
 for i:=1to n do
  a[601-n+i]:=ord(T[I])-48;
 h:=0;
 for i:=601 downto 1 do begin
  a[i]:=a[i]*2+h;
  h:=a[i] div 10;
  a[i]:=a[i] mod 10;
 end;
 for i:=1to 601 do
  if a[i]<>0 then break;
 n:=602-i;
 m:=(n+1)div 2;
 goal(601-m+1);
end.

I think it's OK.
Does anybody know what's wrong with it?

nothing here ;-)

Edited by moderator 09.04.2004 20:56

yixnissac@hotmail.com

What algorithm do you use?
It is necessary to take square root

Thank you for your easy test.
Though it's easy enough,
but it help me find the reason why I crash.
So thanks again