here is my code. please help me :((

#include <stdio.h>
#define HMAX (1<<17)

int N;
unsigned int H[HMAX];

void sift(int x)
{
        int aux, up = x>>1;

        if (x < 1) return;

        if (H[x] > H[up])
        {
           aux = H[x]; H[x] = H[up]; H[up] = aux;
           sift(up);
        }
}

void perc(int x, int n)
{
        int l = x<<1, r = (x<<1)+1, max, aux;

        if (r > n) return;

        max = l;
        if (H[l] < H[r]) max = r;

        if (H[max] > H[x])
        {
           aux = H[max]; H[max] = H[x]; H[x] = aux;
           perc(max, 1+N/2);
        }
}

int main()
{
        int i, val;
        long double sol;
        unsigned int last, blast;

//        freopen("1306.in", "r", stdin);
        scanf("%d", &N);

        H[0] = 4294967295;

        for (i = 1; i <= 1+(N/2); i++)
                scanf("%lu", H+i);
        for (i = 1; i <= 1+(N/2); i++) sift(i);

        for (i = 2+(N/2); i <= N; i++)
        {
                scanf("%d", &val);
                if (val < H[1]) { H[1] = val; perc(1, 1+(N/2)); }
        }


        blast = H[1];
        H[1] = 0;
        perc(1, (N/2));
        last = H[1];

        sol = last*0.5;
        sol += blast*0.5;
        if ( (N&1) == 0 ) printf("%0.1llf\n", sol);
        else printf("%u\n", blast);

        return 0;

}

from the betting coefficient k is (m + n) / n
I think k must be greater  than 1.

Don't you think so?

What is test 6?

Edited by author 11.01.2013 21:21

Can somebody give test #16?
Getting TLE
Thnx

when you search x and y (A*x+B*y = N)
use something like
this for x:= 0 to min(n div a, trunc(sqrt(n))) do
=)

if stones are sorted then answer is n-1

It took me a while to find out this special case :)

this is very bad problem :(

Try the following test:
50
3081 30488 b
16217 39416 w
7248 11364 w
17676 32492 w
25292 30974 w
1686 28199 w
9576 25807 w
18823 21457 w
30775 38642 w
7758 15847 b
6289 19805 b
1546 26246 w
27602 30148 w
9296 14836 w
19435 26471 w
27882 35227 w
24453 42158 w
943 1752 b
4080 12268 w
31529 41419 w
6238 22445 w
6262 13797 w
16616 31697 w
9990 12426 w
31675 46085 w
15149 32698 b
16029 18700 b
32372 50450 w
19642 25152 b
21096 30699 b
27244 35134 b
29951 43374 w
8869 9499 w
3375 28256 w
2060 14099 b
19972 37822 b
30222 33297 b
5433 9482 w
28735 54469 b
19053 31004 w
12234 27889 b
19117 19750 b
11927 17069 w
30056 32152 b
21756 25680 w
25656 44214 b
5308 18661 b
21710 26158 b
9196 32476 b
9917 11034 b

I don't know what the output is. Mine was 54496 1000000000

Nice to see, but "higher" relation is not transitive ;-)

can anybody explain me how to use reals with big range like 10^18 or 10^19 more precisely in pascal?

Please help
Why am I getting floating-point invalid operation on test 3?
Would it be possible to see the test values?

program pr_1;
var n, i, x: longint;
a: real;
begin
read(n);
for i:=1 to n do
begin
  read(x);
  a:= (1 + sqrt(8*x-7))/2;
  if (Frac(a) = 0.0) then
    write('1 ')
  else
    write('0 ');
end;
end.

program acm_1130;

type vec=record
x,y:longint;
end;

var w,a:array[1..3]of vec;
    c,b:vec;
    i,n,nn:longint;
    L:longint;
    p:array[0..10000,1..3,1..2] of byte;
    q:array[0..10000,1..3]of shortint;
    f:boolean;

procedure delta(nn,r:longint);
var j:longint;
begin
 if nn=0 then exit;
 for j:=1 to 2 do
 if p[nn,r,j]>0 then
 begin
  q[nn-1,p[nn,r,j]]:=-q[nn-1,p[nn,r,j]];
  delta(nn-1,p[nn,r,j]);
 end;
end;

procedure minn(i:longint);
begin
  if sqr(a[1].x+a[2].x)+sqr(a[1].y+a[2].y)<=sqr(L) then
  begin
   w[1].x:=a[1].x+a[2].x;
   w[1].y:=a[1].y+a[2].y;
   w[2]:=a[3];
   q[i,1]:=1; q[i,2]:=1; q[i,3]:=1;
   p[i+1,1,1]:=1; p[i+1,1,2]:=2; p[i+1,2,1]:=3;
   exit;
  end;
  if sqr(a[1].x-a[2].x)+sqr(a[1].y-a[2].y)<=sqr(L) then
  begin
   w[1].x:=a[1].x+a[2].x;
   w[1].y:=a[1].y+a[2].y;
   w[2]:=a[3];
   q[i,1]:=1; q[i,2]:=-1; q[i,3]:=1;
   p[i+1,1,1]:=1; p[i+1,1,2]:=2; p[i+1,2,1]:=3; exit;
  end;
  if sqr(a[1].x+a[3].x)+sqr(a[1].y+a[3].y)<=sqr(L) then
  begin
   w[1].x:=a[1].x+a[3].x;
   w[1].y:=a[1].y+a[3].y;
   w[2]:=a[2];
   q[i,1]:=1; q[i,2]:=1; q[i,3]:=1;
   p[i+1,1,1]:=1; p[i+1,1,2]:=3; p[i+1,2,1]:=2; exit;
  end;
  if sqr(a[1].x-a[3].x)+sqr(a[1].y-a[3].y)<=sqr(L) then
  begin
   w[1].x:=a[1].x-a[3].x;
   w[1].y:=a[1].y-a[3].y;
   w[2]:=a[2];
   q[i,1]:=1; q[i,2]:=1; q[i,3]:=-1;
   p[i+1,1,1]:=1; p[i+1,1,2]:=3; p[i+1,2,1]:=2; exit;
  end;
  if sqr(a[2].x+a[3].x)+sqr(a[2].y+a[3].y)<=sqr(L) then
  begin
   w[1].x:=a[2].x+a[3].x;
   w[1].y:=a[2].y+a[3].y;
   w[2]:=a[2];
   q[i,1]:=1; q[i,2]:=1; q[i,3]:=1;
   p[i+1,1,1]:=2; p[i+1,1,2]:=3; p[i+1,2,1]:=1;
   exit;
  end;
  if sqr(a[2].x-a[3].x)+sqr(a[2].y-a[3].y)<=sqr(L) then
  begin
   w[1].x:=a[2].x-a[3].x;
   w[1].y:=a[2].y-a[3].y;
   w[2]:=a[2];
   q[i,1]:=1; q[i,2]:=1; q[i,3]:=-1;
   p[i+1,1,1]:=2; p[i+1,1,2]:=3; p[i+1,2,1]:=1;
   exit;
  end;
end;

begin
  read(n);
  read(L);
  if n=1 then begin write('+'); exit; end;
  if n=2 then
  begin
   read(c.x,c.y);
   read(b.x,b.y);
   if sqr(c.x-b.x)+sqr(c.y-b.y)<sqr(c.x+b.x)+sqr(c.y+b.y) then
   writeln('+-') else writeln('++');
   exit;
  end;
  for i:=1 to 3 do read(a[i].x,a[i].y);
  fillchar(p,sizeof(p),0);
  fillchar(q,sizeof(q),1);

  minn(0); a[1]:=w[1]; a[2]:=w[2];

  for i:=1 to n-3 do
  begin
   read(a[3].x,a[3].y);
   minn(i); a[1]:=w[1]; a[2]:=w[2];
  end;

  f:=true;
  if sqr(a[1].x-a[2].x)+sqr(a[1].y-a[2].y)<sqr(a[1].x+a[2].x)+sqr(a[1].y+a[2].y) then
  f:=true else f:=false;

  if f then
  begin
   nn:=n-2;
   delta(nn,2);
  end;
  writeln('YES');
  for i:=1 to 3 do
  if q[0,i]>0 then write('+') else write('-');
  for i:=1 to n-3 do
  if q[i,3]>0 then write('+') else write('-');

end.

Edited by author 07.01.2013 14:31

What is special about this problem?
I get WA 7 all the time.

My algo
[code deleted]
Edited by author 01.01.2013 16:44

Edited by author 01.01.2013 20:16

Edited by author 01.01.2013 20:18

Edited by moderator 29.11.2019 14:27

#include<iostream>
#include<string>
using namespace std;
int main(){
    int line=1,liner,start=0,paper=1,word,symbol;
    cin>>liner>>symbol>>word;
    for(int i=0;i<word;i++){
            string s;
            cin>>s;
            int a;
            if(start==0) a=s.size();
            else a=s.size()+1;
            if((start+a)<symbol)  start=start+a;
            else {
                 start=0;
                 line++;}
                 }
    if(line%liner==0) cout<<(line/liner);
    else cout<<(line/liner+1);
    system("pause");
    return 0;
}

why the answer to the first sample is 2 I THANK IT IT 1

because let x=1 then f(x)=0 so it only takes 1 sec

That is it? Help

What's wrong with 33th test?))
Or what reazons to access violation?

if N=1 000 000 000 i have Q->555555555888. And as i read it's right answer.

Edited by author 06.01.2013 05:07

how to get a 0.19 in the first test ?
what formula ?

What is it?

Edited by author 05.01.2013 17:57

Edited by author 05.01.2013 17:57