Try to find minimal and maximal distance between figure's center and border, and think about possible shapes.

Who can give me the test 5?I get WA5!Thanks!

help Me please

if WA9, try to check

wwww
wwww
wwwb
wwbb

ans=1

My solution:

program p1060;
type
  frr=array[0..5,0..5] of boolean;
var
  f:frr;
  ans:byte;
procedure init;
  var
    i,j:byte;
    ch:char;
  begin
    fillchar(f,sizeof(f),true);
    for i:=1 to 4 do
      begin
        for j:=1 to 4 do
          begin
            read(ch);
            if ch='b'
              then f[i,j]:=false;
          end;
        readln;
      end;
  end;
function check(f:frr):boolean;
  var
    b:boolean;
    i,j:byte;
  begin
    b:=f[1,1];
    for i:=1 to 4 do
      for j:=1 to 4 do
        if f[i,j]<>b
          then exit(false);
    exit(true);
  end;
function min(a,b:byte):byte;
  begin
    if a<b
      then exit(a)
      else exit(b);
  end;
procedure find(f:frr;x,y,t:byte);
  begin
    if check(f)
      then begin
             ans:=min(ans,t);
             exit;
           end;
    if x>4
      then exit;
    if y=4
      then find(f,x+1,1,t)
      else find(f,x,y+1,t);
    f[x,y]:=not(f[x,y]);
    f[x-1,y]:=not(f[x-1,y]);
    f[x+1,y]:=not(f[x+1,y]);
    f[x,y-1]:=not(f[x,y-1]);
    f[x,y+1]:=not(f[x,y+1]);
    if y=4
      then find(f,x+1,1,t+1)
      else find(f,x,y+1,t+1);
  end;
procedure outit;
  begin
    if ans=17
      then writeln('Impossible')
      else writeln(ans);
  end;
begin
  init;
  ans:=17;
  find(f,1,1,0);
  outit;
end.

Edited by author 07.05.2009 11:48

Edited by author 07.05.2009 11:49

Please, tell me what kind of mistake can I have?
I see that there are a lot of persons
who has problems with Test#12. Is there smth speciale?
Thanks.

Edited by author 14.05.2006 15:27

Try these tests:
(1) - No
10001
11001
10011
11111
11111

(2) - Yes
100001
110001
111001
111011
111111
111111

(3) - Yes
111111
111010
111010
111010
111010
000000

Good luck!

Edited by author 31.07.2010 19:34

9
0 0
1 0
1 1
2 2
3 1
3 0
4 0
4 3
0 3
answer: ccw

Edited by author 31.07.2010 19:05

My submit #3115142 get CE. And report on this submit contains strange symbols.

Edited by author 31.07.2010 01:15

It was really difficult for me, although I read one very good article and understood the idea of the solution. And I don't know how to solve it without any precalculations. I did a lot of them. Does there exists another solution, without using a big recurrence formula?

You can read about this problem here:
http://www.cse.ust.hk/tcsc/RR/2004-02.ps.gz

// 1075.cpp : Defines the entry point for the console application.
//


#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <stdio.h>
using namespace std;
struct point{
    int x;
    int y;
    int z;
    void in(point *);
    double dist(point *,point *);
}Tigran;
void point::in(point *a){
    cin>>a->x;
    cin>>a->y;
    cin>>a->z;
}
double point::dist(point *a, point *b){
    double res_x,res_y,res_z;
    double total_res;
    res_x=((a->x)-(b->x))*((a->x)-(b->x));
    res_y=((a->y)-(b->y))*((a->y)-(b->y));
    res_z=((a->z)-(b->z))*((a->z)-(b->z));
    total_res=sqrt(res_x+res_y+res_z);
    return (total_res);
}
int main(){
    double answer;
    point a,b,c;
    double t;
    double d;
    double x,y,z;
    int R;
    Tigran.in(&a);
    Tigran.in(&b);
    Tigran.in(&c);
    cin>>R;
    x=Tigran.dist(&a,&c);
    y=Tigran.dist(&b,&c);
    z=Tigran.dist(&a,&b);
    if((x*x+z*z<=y*y) && (y*y+z*z<=x*x)){
        answer=z;
    }
    else {
        t=acos((x*x+y*y-z*z)/(2*x*y));
        d=(x*y*sin(t))/z;
        if(d>=(double)R){
            answer=z;
        }
        else {
            answer=sqrt(x*x-R*R)+sqrt(y*y-R*R)+R*(t-acos(R/x)-acos(R/y));
        }
    }
    printf("%0.2f",answer);
    return (0);
}

Задач послано    120
Задач решено    111

Задач решено - ясно как подсчитывается, а вот как считается "Задач послано", как-то понять не могу.
Возможно ли при "задач послано">"задач решено" уравнять эти значения, если да, то как?

Thanks!

is it 312 ?

hi guys
how i must read string, using c#
i mean read the WHOLE string, not just trim before '\n'
sry for my language

that's simply funny - I got result in 1.046sec, but need 1sec... That's unfair..

#include<conio.h>
#include<math.h>
#include<stdio.h>
int k,t[10];
double fonction (int n, int i)
{
       int h;
    if(n==1)
    {
            if(t[n]>=i)
            {
                       return 1;
            }
            else
            {
                return 0;
            }
    }
    else
    {
        if(t[n]>i)
            {
                       return ((pow(10,n-1))+(t[n]*pow(10,n-2))+(fonction(n-1,i)));
            }
        if(t[n]==i)
            {
                       h=int(k/pow(10,n-1));
                       return (((k-(h*pow(10,n-1)))+1)+(t[n]*pow(10,n-2))+(fonction(n-1,i)));
            }
        if(t[n]<i)
            {
                       return ((t[n]*pow(10,n-2))+(fonction(n-1,i)));
            }
    }
}

int main()
{
    int n=0,i=1,j,m,r=1,w=0;
    scanf("%d",&k);
    m=k;
   // double q=1.56;
    while(m!=0)
    {
               n++;
               t[n]=m%10;
               m=m/10;
               //printf("%d",t[n]);


    }
    for(i=0;i<n;i++)
    {
                    w+=pow(10,i);
    }
    r=int(fonction(n,0));
    printf("%d\n",r-w);
    for(i=1;i<10;i++)
    {
    r=int(fonction(n,i));
    printf("%d\n",r);
}
    getch();

return 0;
}

why WA2??
program ural;
var n,a,b,y:byte;
    m,i,z,r:integer;
    h,g:array[1..200]of integer;
    t:array[1..32767]of byte;
    l:array[1..32767]of integer;
    p,q:array[1..32767]of boolean;
    s:array[1..200]of byte;
    k:array[1..400]of integer;
    o:array[1..400]of byte;
    u:array[1..200]of boolean;
    x:boolean;
procedure search(d:byte);
var r:integer;
  begin
    u[d]:=true;
    r:=h[d];
    while r<>0 do
      begin
        if u[t[r]]=false then
          begin
            inc(y);
            k[y]:=g[d];
            g[d]:=y;
            o[y]:=t[r];
            inc(y);
            k[y]:=g[t[r]];
            g[t[r]]:=y;
            o[y]:=d;
            p[r]:=true;
            if q[r] then
              p[r+1]:=true
            else
              p[r-1]:=true;
            search(t[r]);
          end;
        r:=l[r];
      end;
  end;
procedure track(v,d:byte);
var r:byte;
  begin
    if d=i then
      begin
        write(v,' ');
        for r:=1 to v-1 do
          write(s[r],' ');
        writeln(d);
        x:=true;
        exit;
      end;
    u[d]:=true;
    s[v]:=d;
    r:=g[d];
    while r<>0 do
      begin
        if u[o[r]]=false then
          track(v+1,o[r]);
        if x then
          exit;
        r:=k[r];
      end;
  end;
begin
  assign(input,'travel.in');
  assign(output,'travel.out');
  reset(input);
  rewrite(output);
  readln(n,m);
  for i:=1 to m do
    begin
      readln(a,b);
      inc(r);
      l[r]:=h[a];
      h[a]:=r;
      t[r]:=b;
      q[r]:=true;
      inc(r);
      l[r]:=h[b];
      h[b]:=r;
      t[r]:=a;
    end;
  z:=m-n;
  for i:=1 to n do
    if u[i]=false then
      begin
        inc(z);
        search(i);
      end;
  if z<=0 then
    begin
      writeln(0);
      exit;
    end
  else
    writeln(z);
  for i:=1 to n do
    begin
      r:=h[i];
      while r<>0 do
        begin
          if(q[r])and(p[r]=false)then
            begin
              x:=false;
              fillchar(u,sizeof(u),false);
              track(1,t[r]);
            end;
          r:=l[r];
        end;
    end;
end.

Edited by author 29.07.2010 21:50

is there something tricky in it?