Why min base number of the sample is 11? In statement I can't find that k should be >= max digit in an initial number.

BTW if I convert any number to binary it will always be divisible by 1 and as far as I see  will satisfy every condition from the statement (the given number, written in k-based system is divisible by k−1, 2 ≤ k ≤ 36).

Thanks.

Edited by author 13.04.2017 17:00

hey guys,this is my solution.Hope you like it :) C++.


#include <bits/stdc++.h>
using namespace std;
queue<int>qu;                           //queue
#define pii pair<int,int>              //pair
#define maxx 500                      //maximum of anything
vector<pii>vpii[maxx];               //Vector of pair
#define inf 0x7fffffff              //INFINITY
int X[]={+2,+2,-2,-2,+1,-1,-1,+1}; //knight movement left/right
int Y[]={-1,+1,-1,+1,+2,+2,-2,-2};//knight movement up/down
int main()
{
    int pr;
    while(cin>>pr){
         int r=max(0,12-pr);
         int minutes=(4*60)-r*45;
         cout<<(minutes>=0?"YES\n":"NO\n");
    }
    return 0;
}

Edited by author 03.06.2015 10:10

Edited by author 03.06.2015 10:10

main {

prepare_lca();
if (N!=-1) {
 build_segement_tree(1,0,N-2); //a[0..n-1]
}
int Q=read();
while(Q-->0) {
 int j=x;
 int x=read()-1;
 int y=read()-1;
 if (x!=y) {
   J= get(1, 0, N-2, x, y-1);
 }
 println(j);
}

}

in build_segement_tree
..
if (tl==tr) t[v]=lca(tl,tl+1);

vertexes are 1..10  and  d , d + 1, d-1, d+2, d-2,  t + 3, t - 3, where d's are divisors of t.

WA 10
WA 14

My Source

import java.util.*;
public class JavaApplication3 {


    public static void main(String[] args) {
        String SubPalabra1="";
        String SubPalabra2="";
        String Correcta="";

        String Verdadera="";
        int menor;
        String resto="";
        Scanner D=new Scanner(System.in);
        String palabra=D.nextLine().trim();
        if(palabra.length()!=1){
            for(int i=0;i<palabra.length();i++){
                Verdadera="";
                resto=palabra.substring(i+1, palabra.length());
                if(resto.indexOf(palabra.charAt(i))!=-1){
                    SubPalabra1=""+palabra.charAt(i)+resto.substring(0,resto.length());
                      SubPalabra2=""+resto.substring(resto.indexOf(palabra.charAt(i)),resto.length());
                    if(SubPalabra1.length()<SubPalabra2.length()){
                        menor=SubPalabra1.length();
                    }
                    else{
                        menor=SubPalabra2.length();
                    }
                    for(int j=0;j<menor;j++){

                        if(SubPalabra1.charAt(j)==SubPalabra2.charAt(j)){
                            Verdadera=Verdadera+SubPalabra1.charAt(j);
                            if(Verdadera.length()>Correcta.length()){
                                Correcta=Verdadera;
                            }
                        }
                        else{
                            j=menor;
                        }
                    }
                }

            }
        }
        else{
            System.out.println(palabra);
        }
        System.out.println(Correcta);


    }
}

Who is the word that no work?

I have written solution that reached test 22 during the competition(I think it corresponds to 15th in offline mode) but now I have wa and i have run out of ideas. Can anyone give me a clue on this problem. Is there a problem with precision (I am making calculations with long double but use sqrt which is a bit unprecise).

Use queues

KTO NIBUD PONIMAET, KAK EE RESHIT

Do not use "n/3" or "n div 3" or "input div 3" or "input/3".
Do not use any division.
while input n, n:=n*3 ;
and then you won't have to divide at 3 every input;
Sorry for bad english.If any questions please ask.

#include <bits/stdc++.h>
using namespace std;

int main() {
    string letter="abcdefghijklmnopqrstuvwxyz";
    int k=letter.size();
    int i1,i2,i3;
    string part,bigpart;
    for(i1=0;i1<k;i1++)
    {
        for(i2=0;i2<k;i2++)
        {
            for (i3=0;i3<k;i3++)
            {
                part="";
                part.pb(letter[i1]);
                if(part[0]!=letter[i2]) part.pb(letter[i2]); else continue;
                if(part[0]!=letter[i3]&&part[1]!=letter[i3]) part.pb(letter[i3]); else continue;
                bigpart.pb(letter[i1]);bigpart.pb(letter[i2]);bigpart.pb(letter[i3]);
            }
        }
    }
    string ans="";
    //cout<<bigpart.size()/3;
    while(ans.size()<1000000)
    {
        ans=ans+bigpart;
    }
    /*for (int i=0;i<1000000;i++)
    {
        cout<<ans[i];
    }*/
    int res=0;
    for (int i=0;i<1000000;i++)
    {
        if (ans.substr(i,1)=="a") res++;
    }
    cout<<res;
    return 0;
}

Maybe 1 test is a sort of tricky?

Edited by author 06.04.2017 01:18

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

int main()
{
    int N;

    scanf("%d",&N);

    int i,ck = 0;
    for(i = 2; i <= abs(N); i++)
        ck += i;

    if(N < 0)
        printf("-%d\n",ck);
    else
        printf("%d\n",ck);

    return 0;
}

Hello,
Should we multiply by k or n%k if (n - i*k) equals k or n%k ?
Which sample is correct?
Sample 1
8!! = 8 *(8-2)*(8-4)*(8-6)*2

Sample 2
8!! = 8 *(8-2)*(8-4)*(8-6)

any idea what could be my mistake? i'm using
i=n; i< 2n; i++
h2=((mod + h2 - dn * a[i-n] % mod) * d % mod + a[i]) % mod;

where d=256, dn=d^(n-1) mod=10^9+9

2
10
15
16

->
Unrecoverable error. (not "Error in record #1. Correct number is: 1.")

if a route is defined by the stops: 1, 2, 4. Can they go from 4 to 2 then to 1 or is it one-way route?

#include <iostream>
using namespace std;
int main(){
    unsigned int a, b;
    cin >> a >> b;
    cout << a+b;
    return 0;
}

#include <stdio.h>
int main(){
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%d", a + b);
  return 0;
}

import java.util.Scanner;

public class Zadachi {

    public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    int x = 2;
        if(n<=k){ System.out.print(x);
        }else{

            if (n*2 % k == 0){
                System.out.print(n*2/k);
            }else{
                System.out.print(n*2/k+1);
            }
        }

}
}