IN: 3 0
OUT: 0

This test helped me :D
Good luck!

What should I do, if some point O[i] has negative z-coordinate?

Edited by author 11.10.2014 17:37

from were i can get tests for problems or this is unavailable information?

Maxim Buzdalov have added a new bunch of tests. Progbeat, showjim and [NRU ITMO] WiNGeR have lost their AC.

Edited by author 07.10.2014 13:20

import java.util.Scanner;
import java.util.Collections;
import java.util.ArrayList;

public class Timus{
    public static void main(String[] args){

        Scanner in = new Scanner(System.in);
        ArrayList<Integer> arr = new ArrayList<Integer>();
        int n = in.nextInt();
        for (int i = 0; i< n; ++i){
            arr.add(in.nextInt());
        }
        int b = in.nextInt();
        for (int i = 0; i< b; ++i){
            arr.add(in.nextInt());
        }
        b = in.nextInt();
        for (int i = 0; i< b; ++i){
            arr.add(in.nextInt());
        }
        int sum=0;
        int occurrences = 0;
        for(int j =0; j<n; ++j){
            occurrences = Collections.frequency(arr, arr.get(j));
            if (occurrences % 3 == 0){
                sum = sum + occurrences;
            }
        }

        System.out.print(sum/3);


    }
}

Может ли Жаба пролететь в отверстие? Если да, можно ли бросать камешек уже когда жаба пролетела внутрь?

what s wrong this time ?
anybody ?


#include<iostream>
#include<string>

using namespace std;

char napis[1001]={0};

char tab[35]={0};
int tab1[35]={0};

int main(){
tab[0]=*"a"; tab1[0]=1;
tab[1]=*"b"; tab1[1]=2;
tab[2]=*"c"; tab1[2]=3;
tab[3]=*"d"; tab1[3]=1;
tab[4]=*"e"; tab1[4]=2;
tab[5]=*"f"; tab1[5]=3;
tab[6]=*"g"; tab1[6]=1;
tab[7]=*"h"; tab1[7]=2;
tab[8]=*"i"; tab1[8]=3;
tab[9]=*"j"; tab1[9]=1;
tab[10]=*"k"; tab1[10]=2;
tab[11]=*"l"; tab1[11]=3;
tab[12]=*"m"; tab1[12]=1;
tab[13]=*"n"; tab1[13]=2;
tab[14]=*"o"; tab1[14]=3;
tab[15]=*"p"; tab1[15]=1;
tab[16]=*"q"; tab1[16]=2;
tab[17]=*"r"; tab1[17]=3;
tab[18]=*"s"; tab1[18]=1;
tab[19]=*"t"; tab1[19]=2;
tab[20]=*"u"; tab1[20]=3;
tab[21]=*"v"; tab1[21]=1;
tab[22]=*"w"; tab1[22]=2;
tab[23]=*"x"; tab1[23]=3;
tab[24]=*"y"; tab1[24]=1;
tab[25]=*"z"; tab1[25]=2;
tab[26]=*"."; tab1[26]=1;
tab[27]=*","; tab1[27]=2;
tab[28]=*"!"; tab1[28]=3;
tab[29]=*" "; tab1[29]=1;




    int i=0,j=0,k=0,licznik=0;
    cin.getline (napis, 1000);
    k = strlen(napis);
    for(i=k-1;i>=0;i--){
            for(j=0;j<30;j++){
                    if(napis[i]==tab[j]){
                               licznik+=tab1[j];
                               }
                    }
            napis[i]=0;
            }
    cout << licznik;
    getchar();
    return 0;
}

This problem solved with the left edge.

We look for position of "2".

We have three cases:
1) 12***** - "2" in the second position
2) 1*2**** - "2" in the third position

    1**2*** - "2" in the fourth position is impossible
    1***2** - "2" in the fifth position is impossible

3) 1******2 - "2" in the last position is possible (the one special case).
For example: 1357642

For each case find sub-task:
1) a[i-1]
2) is NOT a[i-2]
For 1*2**** there can only be one 132****, more accurate 1324***.
Therefore is a[i-3].
3) is 1

Recurrent expression: a[i] = a[i-1] + a[i-3] + 1.
And a precalculation:
a[1] = 1
a[2] = 1
a[3] = 2

A some answers to compare:
a[4] = 4
a[5] = 6
a[6] = 9
a[7] = 14

what is the difference?? to count as whole stakes or to count it with side.
as i think this is the same problem but we get different answers to this question.

Who knows when the contest problems will be avaible in the Archieve?

http://acm.timus.ru/getsubmit.aspx/5892732.cpp

Это решение содержит кучу assert-ов на то, что число операций не более 10^6, но все равно падает по TL на 9-м тесте. Что это?

Если 2 человека пытаются заплатить за квартиру нечётной стоимости, что происходит?

что ели кто то наример я захочу добавить задачу в архив что необходимо для этого делать?

I tried everything to do that but i got TLE, and then MLE.

So I used index table with 1024*1024 elements and got AC.
But the problem said that 10^9.

Maybe it's a joke ?

Hello problem is to find the summ of all k-based numbers?

Edited by author 30.04.2014 12:54

Find the solution and explanation of this task in
manyprogrammingtutorials.blogspot.com

program sdf;
var n,n1,n2,n3,i,i1,i2,k,a:integer;
m,m1,m2,e:array[1..1000 ] of integer;
begin
    k:=1;
readln(n1);
For i:=1 to n1 do readln(m[i]);
readln(n2);
For i:=1 to n2 do readln(m1[i]);
readln(n3);
For i:=1 to n3 do readln(m2[i]);
For i:=1 to n1 do
for i1:=1 to n2 do
for i2:=1 to n3 do begin
if (m[i]=m1[i1]) and (m1[i1]=m2[i2]) and (m[i]+m1[i1]+m[i2]<>-3) Then begin
                                          e[k]:=m[i];
                                          k:=k+1;
                                          m[i]:=-1;
                                             m1[i1]:=-1;
                                             m2[i2]:=-1;
                                             end;
 end;


For i:=1 to k do begin
for i1:=i+1 to k do begin
if e[i]=e[i1] Then e[i1]:=-1;
end;
end;
For i:=1 to k-1 do begin
if e[i]<>-1 Then a:=a+1;
end;
writeln(a);
end.
 в ЧЕМ ошибка?

What is test 3??? Help me!!! Please!!!

    int victim;
    while (scanf("%d", &victim) != EOF)
    {
        /*do something*/
    }
it is an infinite loop