Notice that the number of pieces you cut the cables needn't be precisely equal to k. Just >=k is OK.

Set types as long long for variables used as sum of cables lengths

Try this:
1 5
1.00
answer: 0.20

Edited by author 02.04.2021 18:58

if you read data as:
double x;
cin>>x;
a[i]=x*100;
it doesn't work!

In test #4 n=10000 and k=1.
You can write this case apart.

program Project1;

{$APPTYPE CONSOLE}

uses
  SysUtils;

   var
   i, j, m, n, maxa, x0, xn, x, p : integer;
   s : int64;
   l : real;
   a : array [1..10000] of integer;

begin
   readLn(n, m);
   maxa := 0;    s := 0;
   for i := 1 to n do
     begin
       readLn(l);
       a[i] := round(100*l);
       s := s + a[i];
       if maxa < a[i] then maxa := a[i];
     end;
   if s div m < 1 then writeLn('0.00')
   else
     begin
       if m = 1 then writeLn(maxa/100:10:2)
       else
         begin
           xn := s div m;
           x0 := 0;
           repeat
             p := 0;
             x := (xn+x0) div 3;
             for i := 1 to n do
               inc(p, a[i] div x);
             if p >= m then x0 := x
             else xn := x;
           until p = m;
           for i := xn downto x0 do
             begin
               p := 0;
               for j := 1 to n do
               inc(p, a[j] div i);
               if p = m then begin writeLn(i/100:0:2);  halt;  end;
             end;
         end;
     end;
end.

I use binary search to solve this problem and had trouble
with input
this input gots AC:
   double x;
   int j,t;
        fscanf(inf,"%lf",&x);
        L[i]=(int)(x*100.0+1e-9);
but this got WA9
        fscanf(inf,"%i.%i",&t,&j);
        L[i]=t*100+j;
 output in boyh programs was like this
    fprintf(outf,"%i.%0.2i\n",ANS/100,ANS%100);

possible reasons:
1. input data hasn't 2 digits after decimal point(maybe it have none digits and no point at all)
2. chacker is floating-point written program with do errors
and you have to do the same errors to AC.

#include <stdio.h>
#include <math.h>

int n, k;
double * data;

int ok (double z) {
 long s = 0;
 for (int i = 0; i < n; i++)
  s += floor (data[i] / z);
 return s >= k;
}

int main () {
 scanf ("%d%d", &n, &k);
 data = new double [n];
 double l;
 for (int i = 0; i < n; i++) {
  scanf ("%lf", &l);
  data[i] = l;
 }
 if (!ok(0.01)) {
  printf ("0.00\n");
  return 0;
 }
 double x = 0.01, y = 1e5, z;
 while (y - x >= 0.0001) {
  z = (x + y) / 2;
  if (ok(z)) x = z;
  else y = z;
 }
 printf ("%.2lf\n", z);
 return 0;
}

I read every line containing a number. And after that simply read all digits (without a point) and convert into integer 1 cm=1. 100km=100*10^3m=100*10^5cm=10^7. In that range I use binary search for finding the maximum possible length. Why WA at test 1?  (Yeah, I sort the lens too)
Here is a sample of code for converting:
fgets(buf,150,INPUT_STREAM);
       tmp=0;
       j=0;
       while((buf[j]>='0'&&buf[j]<='9')||buf[j]=='.')
      {
         if(buf[j]!='.')tmp=tmp*10+buf[j]-'0';
         j++;
      }
      sum+=tmp;
      Lens[i]=tmp;

Do the lengths always have exactly two digits

Edited by author 22.08.2006 13:44

my solution:

{$N+}
program cable_master;
const max=10000;
var a:array[1..max] of real;
    n,k,s:integer;
    best,f,r,m:extended;
procedure readdata;
var i:integer;
begin
readln(n,k);
for i:=1 to n do
 begin
  readln(a[i]);
  if a[i]>r then r:=a[i];
 end;
end;
function find:boolean;
var i:integer;
begin
s:=0;
for i:=1 to n do
  s:=s+trunc(a[i]/m);
if s=k then find:=true
 else find:=false;
end;
procedure main;
begin
while r>=f do
 begin
  m:=(f+r)/2;
  if not find then
    begin
     if s>k then f:=f+0.001
      else r:=m;
     if abs(r-f)<0.00001 then break;
    end
   else
    begin
     if m>best then best:=m;
     if f=r then break;
     f:=f+0.001;
    end;
 end;
best:=trunc(best*100);
best:=best/100;
writeln(best:0:2);
end;
BEGIN
readdata;
main;
END.

// My algorithm is about binary search

#include <stdio.h>
#include <stdlib.h>
#define maxd 10005
float d[maxd],m=0.0;
long k,n;
int check(float w)
{
    long l,s=0;
    for (l=0; l<n; l++)
    {
        if (s>=k) return 1;
        if (w>d[l]) return -1;
        s+=(long)(d[l]/w);
    }
    if (s>=k) return 1;
    return -1;
}
int sort_function(const void*a,const void*b)
{
    if ((*(float*)a)>(*(float*)b)) return -1;
    if ((*(float*)a)==(*(float*)b)) return 0;
    return 1;
}
int main()
{
    long l;
    float a,b,c,tmp;
    FILE* fname=stdin;
    FILE* fout=stdout;
    fscanf(fname,"%ld%ld",&n,&k);
    for (l=0; l<n; l++) fscanf(fname,"%f",&d[l]);
    qsort((void*)d,n,sizeof(d[0]),sort_function);
    c=(float)((long)((long)100*(long)(1+10000)/2))/100.0;
    for (a=1,b=10000; a<=b; c=(float)(a+b)/2.0 )
    {
        if (check(c)==1)
        { if (c>m)
            m=c;
            tmp=a;
            a=c+0.01;
            a=(float)((long)
(100.0*a))/100.0;
            if (a==tmp) break;
        }
        else
        {
            b=c-0.01;
            b=(float)((long)
(100.0*b))/100.0;
        }
    }
    m=((long)(m*100))/100.0;
    fprintf(fout,"%.2f\n",m);
    fclose(fname); fclose(fout);
    return 0;
}

Here it is:

#include<iostream.h>
#include<stdio.h>

int  n, k;
long x[10000];

void main()
{
double xx;
int i, j;
    cin>>n>>k;
    for (i=0; i<n; i++)
    {
        cin>>xx;
        x[i] = xx*100;
    }
long l, r, m, w;
    r = x[0];
    for (i=1; i<n; i++)
        if (x[i]>r) r=x[i];
    l = 1;
    w = 0;
    while ( l <= r )
    {
        m = (l + r) / 2;
        j = 0;
        for (i=0; i<n; i++)
              j = j + (long)x[i]/m;
        if ( j == k )
        {
            w = m;
        }
        if (j >= k)
            l = m + 1;
        else
            r = m - 1;
    }
double ans = (double)w/100.0;
    printf("%.2f\n",ans);
}

#include <iostream.h>

int n,k;
long m [10000],how [10000];

long GetNewLength ()
{
    long NewLength = 0;
    for (int i=1; i<=n; i++)
        if (m[i]/(how[i]+1) > NewLength) NewLength = m[i]/(how
[i]+1);
    return NewLength;
}

int main ()
{
    long nowLength = 2147483648, nowN = 0;
    double length;
    cin >> n >> k;
    for (int i=1; i<=n; i++)
    {
        cin >> length;
        m [i] = int (length * 100);
        if (m[i] < nowLength) nowLength = m [i];
    }
    while (nowN<k && nowLength != 0)
    {
        nowLength = GetNewLength ();
        if (!nowLength) continue;
        nowN = 0;
        for (i=1; i<=n; i++)
        {
            how [i] = m [i] / nowLength;
            nowN = nowN + how [i];
        }
    }
    cout << nowLength/100 << ".";
    int tens = nowLength%100;
    if (tens < 10) cout << "0" << tens;
        else cout << tens;
    return 0;
}

What could be wrong with this code?

Program Easy;
 Const Max=10006;
 Var i,n,k:Longint;
     a:Array[1..Max] Of Longint;
     x,y,z:Longint;
     u,q:Real;
 Function ok(l:Longint):Boolean;
  Var i,s:Longint;
 Begin
  s:=0;
  For i:=1 To n Do
   s:=s+a[i] Div l;
  ok:=s>=k;
 End;
Begin
{ Assign(Input,'1184.in'); Reset(input);{}
 Read(n,k);
 y:=10000000;
 For i:=1 To n Do Begin
  Read(u);
  a[i]:=Round(u*100);
 End;
 x:=1;
 If Not ok(x) Then Begin
  Writeln('0.00');
  Halt;
 End;
 While y-x>1 Do Begin
  z:=(y+x) Div 2;
  If ok(z) Then
   x:=z
  Else
   y:=z;
 End;
 While Not ok(z) Do
  z:=z-1;
 While Ok(z+1) Do
  z:=z+1;
 q:=z;
 q:=q/100;
 Writeln(q:0:2);
End.

Here is my programme:

const    max    = 10000;

var      a                :array[1..max] of longint;

         n,k              :word;

         dau,cuoi,giua,
         kq               :longint;

         ok               :boolean;


procedure nhap;
var    i       :word;
       x       :real;
begin
  readln( n, k);

  for i := 1 to n do
  begin  readln( x);
         a[i] := trunc( x * 100);
  end;
end;


procedure cut;
var    doan     :longint;
       i        :word;
begin
  doan := 0;
  for i := 1 to n do
  begin
    doan := doan + a[i] div giua ;
    if doan >= k then break;
  end;

  ok :=  ( doan >= k );
  if ok and ( giua > kq ) then kq := giua;
end;


procedure solve;
begin
  dau := 1; cuoi := 10000000; kq := 0;

  repeat
    giua := ( dau + cuoi ) div 2;

    cut;

    if ok then dau  := giua + 1
          else cuoi := giua - 1;
  until dau > cuoi;
end;


procedure out;
var     x     :real;
begin
  x := kq;
  x := x / 100;
  writeln(x :0 :2);
end;


begin
  nhap;
  solve;
  out;
end.