don't round up. or you will get wa8.



Edited by author 28.05.2023 19:05

Guess, why I had wasted 18 attempts?
The exponent can have + sign: 1.3e+32

All bad wished to author.

I was trying C++ for several hours.  Can't defeat test case No.  7. So,  I gave up and got AC using Python Decimal module just after several minutes I stopped trying C++.  It is bad I think that I don't solved with C++.  Literally no idea why WA#7. I tried rounding and truncation.  The result was always the same.

This is my programm below:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Globalization;

namespace _1248
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            double s = 0;
            for (int i = 0; i < n; i++)
            {
                string strS = Console.ReadLine();
                double ch =0;
                if (strS.Contains("e"))
                {
                    string[] arr = strS.Split('e');
                    double ss = double.Parse (arr[1]);
                    ch = double.Parse(arr[0], CultureInfo.InvariantCulture) *  Math.Pow(10, ss);
                }
                else
                    ch = double.Parse  (strS );

                s +=ch;
            }

            StringBuilder sb = new StringBuilder();
            sb.AppendFormat("{0:0.000000000000000000e0}", s);
            string sss = sb.ToString();
            sss = sss.Replace(',', '.');
            //sss = sss.Replace("e0", "");
            Console.WriteLine(sss);
        }
    }
}

I tried everything before I realized that my program should accept numbers like 1.e+1 even if there is no digits after '.'. Weird! I hope this information would be helpful to someone.

Почему моя программа выдаёт неправельный ответ на 6 вопрос?

8
9.23e4
1.23e0
12.2e3
13.233e1
12.2e-3
9.2e-1
10.22e-3
123.12e-1


92300
1.23
12200
132.33
0.0122
0.92
0.01022
12.312
----------------------  +
104646.81442

Result: 1.046468144200000000e5


Remember the "." doesn't have to be
eg.

9e0
9e2
9e-2


Also you don't have to round your result! i have AC

Edited by author 19.04.2011 20:29

Edited by author 19.04.2011 20:31

just ACed.... spent two hours debugging.

very nice problem!

Edited by author 02.12.2011 05:00

If you get wrong answer on test #6,you should notice that there may be a '+' following the 'e'.

when the answer is 0, what's the output?


Edited by author 19.11.2010 20:00

Because I got AC without any trunc or round!

Edited by author 15.11.2010 17:38

Edited by author 09.11.2010 00:31

Checker accepts any number of digits in mantissa, so it is possible to output just exact answer having all digits. Rounding to 19 digits or appending zeros to 19 digits is not required.

For answer calculated in java.math.BigDecimal the following simple statement works:
out.println(sum.toPlainString() + "e0");

{$N+}
program a1;
var i,j,n,pl:word;
    s,sum:Extended;
    res:string;
    fl,fl2:boolean;
begin
read(n);
read(s);
sum:=s;
for i:=1 to n-1 do
  begin
    read(s);
    sum:=sum+s;
  end;
str(sum,res);
n:=length(res);
fl:=false;
fl2:=true;
for i:=1 to n do
  begin
    if res[i]='E' then
      begin
        res[i]:='e';
        pl:=i;
        fl:=true;
      end;
    if (fl=true){ and (fl2=true)} and (res[pl+1]='0') or (res[i]='+') then
      begin
        i:=i-1;
        fl2:=false;
        for j:=pl+1 to n do
          begin
            res[j]:=res[j+1];
            if res[j]<>'0' then fl2:=false;
          end;
      end;
  end;
for i:=1 to length(res)-1 do res[i]:=res[i+1];
write(res);

end.

Please give me test 6. Thanks

Help me, PLEASE!!!!!!!!!!!!!!!!!!!!!

 var
a,b:array[-250..250] of byte;
x1,x,i,j,n,code:integer; s:string;
procedure E;
  var i,ymn,t:integer; minus:boolean;
begin
  minus:=false; ymn:=0;  t:=0;
  for i:=1 to length(s) do
  if (s[i]='e') or (s[i]='E') then
  begin
   t:=i+1;
   val(copy(s,t,length(s)-t+1),ymn,code);
   delete(s,i,length(s)-i+1);
   break;
  end;
  t:=0;
  for i:=1 to length(s) do
  if s[i]='.' then t:=i-1;
  if t<>0 then delete(s,t+1,1);
  if t=0 then t:=length(s);
  ymn:=ymn+t;
  if ymn<=0 then
   begin
    for i:=1 to abs(ymn) do s:='0'+s;
    s:='.'+s;
    s:='0'+s;
   end
  else
   begin
    if ymn<length(s) then insert('.',s,ymn+1)
    else
    begin
       ymn:=ymn-length(s);
       for i:=1 to ymn do s:=s+'0';
    end;
   end;
end;
begin

{   assign(input,'c:\test.txt');
   reset(input);}
   readln(n);
   for i:=1 to n do
      begin
         fillchar(b,sizeof(b),0);
         readln(s);
         if pos('.',s)=0 then
         if pos('e',s)=0 then s:=s+'.0' else insert('.0',s,pos('e',s));
         E;
         if pos('.',s)=0 then s:=s+'.0';
         for j:=pos('.',s)-1 downto 1 do begin val(s[j],x,code); b[j-pos('.',s)]:=x; end;

         for j:=pos('.',s)+1 to length(s) do begin val(s[j],x,code); b[j-pos('.',s)]:=x; end;

         x:=0;

         for j:=200 downto 1 do
           begin
                 x1:=x;
                 x:=(b[j]+a[j]+x1) div 10;
                 a[j]:=(a[j]+b[j]+x1) mod 10;
           end;

         for j:=-1 downto -200 do
               begin
                 x1:=x;
                 x:=(b[j]+a[j]+x1) div 10;
                 a[j]:=(a[j]+b[j]+x1) mod 10;
               end;
      end;
{      for i:=-10 to 10 do write(a[i]);}
    x:=0;
    for i:=-200 to 0 do if a[i]<>0 then break;
    if i<>0 then
      begin
         write(a[i],'.');
         x:=i+1;
         j:=0;
         for i:=x to 200 do
         if i<>0 then
            begin
              inc(j);
              if j=19 then break else write(a[i]);
            end;
         write('e',abs(x));
      end
    else
    begin
    j:=0;
    for i:=1 to 200 do if a[i]<>0 then break;
    x:=-(i);
    if i=200 then x:=0;
    write(a[i],'.');
    for i:=abs(x)+1 to 250 do
      begin
              inc(j);
              if j=19 then break else write(a[i]);
      end;
     write('e',x);
    end;
end.

I've got AC using long number arithmetics. Somebody solve this problem without the use of long arithmetic?

21
9e0
9e1
9e2
9e3
9e4
9e5
9e6
9e7
9e8
9e9
9e10
9e11
9e12
9e13
9e14
9e15
9e16
9e17
9e18
9e19
9e20

Correct answer is
1.000000000000000000e21