#define I(x) int x; cin >> x
#define F(i,n) for(int i = 0; i < (n); ++i)
int main(){
  cin.sync_with_stdio(false);
  I(n); I(k); I(m); cout << "YES" << endl;
  F(i,k){
    I(s); F(i,s){ I(x); //...

1. Инициализация
Начинайте с 1 конюшни
Пусть нам надо поставить туда I лошадей (I = 1..N)
Коэффициент несчастья для I лошадей будет при этом равен White * Black для Horses = 0..i
По итогу этого у нас будут оптимальные (потому что единственные) коэффициенты для всех лошадей в случае 1 конюшни

2. Итерации (меняем J-ую конюшню)
Надо рассчитать минимальный коэффициент несчастья если в 1..J конюшни поставить I лошадей
Для этого переберём все варианты, поставив в J конюшню K лошадей (K = 1..I лошади) и оптимально заполнив первые J-1 конюшни I-K лошадьми (этот коэффициент у нас уже рассчитан на предыдущем шаге итерации)
Перебрав все варианты у нас получится оптимальное заполнение первых J конюшен первыми I лошадьми
Переходим к J + 1 конюшне и считаем заново все варианты заполнения уже для неё

Hi, I had an accepted solution 3 years ago and now I have no idea what was wrong/right. Looking at my previous solution that now gets TLE, it was Kuhn matching algorithm. Did the constraints change or did you just add better tests?

I found my mistake. I have AC.

Edited by author 12.12.2010 15:33

worst statement in timus

I got AC with

#define I(x) int x; cin >> x
#define F(i,n) for(int i = 0; i < (n); ++i)
int main(){
  cin.sync_with_stdio(false);
  I(n); I(m);
  F(i,n){
    string w; cin >> w; I(j);
    auto& c = clones[j-1];
    if(w == "learn"){
    // ...

For wa#3:
4 0
0 3
0 0
YES

For wa#4
0 0
5 0
6 5
YES

Could someone help me ?

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    double outputNumbers[100000];
    double number;
    int index;

    index = -1;
    while (cin.fail() == false)
    {
        cin >> outputNumbers[++index];
    }

    cout << fixed;
    cout << setprecision(4);

    while (index > 0)
    {
        cout << sqrt(outputNumbers[--index]) << endl;
    }

    return 0;
}

My program:

Program t1127;

Const P:array[1..24,1..6]of integer= {all combinations}
       ( (6,4,3,1,5,2),
         (3,5,4,1,6,2),
         (4,6,5,1,3,2),
         (5,3,6,1,4,2),
         (6,4,5,2,3,1),
         (3,5,6,2,4,1),
         (4,6,3,2,5,1),
         (5,3,4,2,6,1),
         (1,2,6,3,4,5),
         (6,4,2,3,1,5),
         (2,1,4,3,6,5),
         (4,6,1,3,2,5),
         (1,2,4,5,6,3),
         (6,4,1,5,2,3),
         (2,1,6,5,4,3),
         (4,6,2,5,1,3),
         (1,2,3,4,5,6),
         (3,5,2,4,1,6),
         (2,1,5,4,3,6),
         (5,3,1,4,2,6),
         (1,2,5,6,3,4),
         (3,5,1,6,2,4),
         (2,1,3,6,5,4),
         (5,3,2,6,1,4)  );
   MaxN=1000;

Var Cube      :array[1..MaxN,1..6]of char;
    Tmp       :array[1..6]of char;
    ACube     :array[1..MaxN,1..24]of string[4];
    yet       :array[1..MaxN,1..24]of boolean;
    N,i,j,k   :integer;
    max,ik,jk :integer;
    m         :integer;
    ch        :char;

begin
Read(n);
for i:=1 to N do
 for j:=1 to 6 do begin
  read(ch);
  while (ch=#10)or(ch=#13)or(ch=#32) do read(ch);
  Cube[i,j]:=ch;
 end;
for i:=1 to N do
 for j:=1 to 24 do begin
  for k:=1 to 6 do tmp[k]:=cube[i,p[j,k]];
  ACube[i,j]:='';
  ACube[i,j]:=tmp[1]+tmp[2]+tmp[3]+tmp[4];
 end;
m:=0;
for i:=1 to N do
 for j:=1 to 24 do
  yet[i,j]:=true;
for i:=1 to N do
 for j:=1 to 24 do if yet[i,j] then begin
  yet[i,j]:=false;
  max:=1;
  for ik:=1 to N do if ik<>i then
   for jk:=1 to 24 do
    if ((acube[i,j][1]=acube[ik,jk][1])and(acube[i,j][4]=acube[ik,jk]
[4]))or
       ((acube[i,j][1]=acube[ik,jk][4])and(acube[i,j][4]=acube[ik,jk]
[1])) then
    if ((acube[i,j][2]=acube[ik,jk][2])and(acube[i,j][3]=acube[ik,jk]
[3]))or
       ((acube[i,j][2]=acube[ik,jk][3])and(acube[i,j][3]=acube[ik,jk]
[2]))
       then begin
     yet[ik,jk]:=false;
     max:=max+1;
     break;
    end;
  if max>m then m:=max;
 end;
writeln(m);
end.

import java.util.Scanner;
 public class Problem_1881 {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int h,w,n,i,p,ans,SL,h1;
        ans=SL=h1=0; p=-1;
    .....

Edited by author 22.03.2014 20:32

Can you give me a test similar to the fifth test. Just i dont know where i have endless cycle? Sorry for my english.

Why?
My code works correctly when it is no "end" in the end and with !=

qayerda xato qildim

What is wrong with mycode? it is working in mycomputer but failing in test2

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
/**
 * Created by sherxon on 3/17/14.
 */
public class Biathlon {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n=Integer.parseInt(in.nextLine());
        String s, name, time;
        Double min, sec;
        int j=0;
        List<String> a= new ArrayList();
        ArrayList<Double> b= new ArrayList();
        int thirty=0;

           for (int i = 0; i <n ; i++) {

               s = in.nextLine();


            name=s.split(" ")[0];

            sec=Double.parseDouble(s.split(" ")[1].split(":")[1])+60*(Double.parseDouble(s.split(" ")[1].split(":")[0]));
               sec+=thirty;
               //System.out.println(sec);
               double eachsec=0.0;
               double eachsecs=0.0;
                boolean bl=true;


               for (int k = 0; k<a.size() ; k++) {
                 //  System.out.println(sec-thirty + " " + (b.get(k)-eachsec) + " " +(sec) + " " + b.get(k));


                   if(sec-thirty<b.get(k)-eachsec&&sec<b.get(k)){
                       b.remove(k);
                       a.remove(k);
                       k--;
                   }
                   eachsec+=30;
               }
               for (int k = 0; k <a.size() ; k++) {
                   if(sec-thirty>b.get(k)-eachsecs){
                       bl=false;
                   }
                   eachsecs+=30;
               }
                if(bl){
               a.add(name);
               b.add(sec);
                }

               thirty+=30;
        }
        Collections.sort(a);
        System.out.println(a.size());

        for (int i = 0; i <a.size() ; i++) {
            System.out.println(a.get(i));
        }
    }
}

Can variables have names "or", "and", "xor", "not", "print", "if", "goto", "end"?

Why for test 1 0 1 0 3 1 0 0 answer
EF-
AE-
DH+
CD-
EH-
is wrong?

Could someone publish such testcase with the answer = -1?

So I've made an embarrassingly large number of submissions for this problem. Thing is I submitted in C++ 11 and got WA 1. Afterwards I switched to Visual C++/G++ and got WA 2. Apparently this was caused by asserts that I had left in my program to check if test 1 is actually the example. After I removed them I got Accepted, but still WA 1 in C++11. So my questions would be:

1. Why did it get WA in C++ 11?
2. Why didn't the assert crash the program?

import java.util.Scanner;

public class BicycleCode{
    public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int firstCode = sc.nextInt();
    int secondeCode = sc.nextInt();

    if(firstCode%2==0 && secondeCode%2!=0){
    System.out.println("yes");
    }else if(firstCode%2!=0 && secondeCode%2==0){
    System.out.println("no");
    }
    }
}