#include <iostream>
long int a,b,c,d,e,f,n,z,x,a1,a2;
using namespace std;
void main() {
    cin >>a;
    n=a%10;
    f=a%100;
    f=f/10;
    e=a%1000;
    e=e/100;
    d=a%10000;
    d=d/1000;
    c=a%100000;
    c=c/10000;
    b=a%1000000;
    b=b/100000;
    if ((n==9)||(f==9)||(e==9)) {
        z=(e+f+n)+1%10;
    } else {
        z=e+f+n;
    } if ((d==9)||(c==9)) {
        x=(b+c+d)+1%10;
    } else {
        x=b+c+d;
    }
    a1=z-x;
    a2=x-z;
    if ((a1==1)||(a2==1)) {
        cout <<"Yes";
    } else {
        cout <<"No";
    }
}

4
0 0
1 0
0 1
-1 1

0.7071


3
0 0
1 0
0 1

0.6436


3
-1 0
1 0
0 1

0.9102

Test#1
3 3
1 2 3
ans:27

Test#2
2 6
1 3
ans:36

Test#3
3 6
3 4 5
ans: 180

Test#4
1 5
3
ans:5

Test#5
1 2
1
ans:1

Test#6
3 2
1 2 2
ans:4

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Timus2002 {

    public static void main(String[] args) {

        Map<String, String> map = new HashMap<>();
        Scanner sc = new Scanner(System.in);
        String[] s = new String[105];
        ArrayList<String> logged = new ArrayList<>();
        int n = Integer.parseInt(sc.nextLine());
        for (int i = 0; i < n; i++) {
            s[i] = sc.nextLine();
        }

        for (int i = 0; i < n; i++) {
            String[] mas = s[i].split(" ");
            if(mas[0].equals("register")){
                if(map.containsKey(mas[1])){
                    System.out.println("fail: user already exists");
                    continue;
                }
                map.put(mas[1], mas[2]);
                System.out.println("success: new user added");
            }

            String login = mas[1];
            if(mas[0].equals("login")){
                String pass = mas[2];
                if(!map.containsKey(login)) {
                    System.out.println("fail: no such user");
                    continue;
                }
                String value = map.get(login);
                if(!value.equals(pass))
                    System.out.println("fail: incorrect password");

                if(value.equals(pass)) {
                    if (!logged.contains(login)) {
                        logged.add(login);
                        System.out.println("success: user logged in");
                    }
                    else
                        System.out.println("fail: already logged in");

                }
            }

            if(mas[0].equals("logout")) {
                if (!map.containsKey(login))
                    System.out.println("fail: no such user");
                if (!logged.contains(login))
                    System.out.println("fail: already logged out");
                else {
                    System.out.println("success: user logged out");
                    logged.remove(login);
                }
            }
        }
    }
}

Pleae help. what is this test?
Any input i get correct answer in my compiler.But here i get WA #10

Try this

1 1
2 2
3 3
4 4

2 change

how can i see my current accepted submission rank????

#include<iostream>
using namespace std;
int n,i,k,t,j;
string a[10000],b[10000],c[10000],d[10000];
int main(){
        a[1]="Alice"; a[4]="Phil";  a[7]="Phoebus";
        a[2]="Ariel"; a[5]="Peter";  a[8]="Ralph";
        a[3]="Aurora"; a[6]="Olaf";     a[9]="Robin";

        b[1]="Bambi"; b[4]="Mulan";  b[7]="Silver";
        b[2]="Belle"; b[5]="Mowgli";  b[8]="Simba";
        b[3]="Bolt";  b[6]="Mickey";  b[9]="Stitch";

        c[1]="Dumbo"; c[4]="Kuzko"; c[7]="Tarzan";
        c[2]="Genie"; c[5]="Kida";  c[8]="Tiana";
        c[3]="Jiminy"; c[6]="Kenai"; c[9]="Winnie";
        cin>>n;
            for(i=1; i<=n; i++) cin>>d[i];
                for(i=1; i<=n; i++) {
                    for(j=1; j<=9; j++)  if(d[i]==a[j]){
                    if(t==0) k+=0;
                    if(t==1) k+=1;
                    if(t==2) k+=1;
                    if(t==3) k+=2;
                    t=1;
                    break;
                    }
                    for(j=1; j<=9; j++)  if(d[i]==b[j]){
                    if(t==0) k+=0;
                    if(t==1) k+=1;
                    if(t==2) k+=1;
                    if(t==3) k+=1;
                    t=2;
                    break;
                    }
                    for(j=1; j<=9; j++) if(d[i]==c[j]) {
                    if(t==0) k+=0;
                    if(t==1) k+=2;
                    if(t==2) k+=1;
                    if(t==3) k+=1;
                     t=3;
                     break;
                     }
}
            cout<<k;
}

what's wrong with this code:
#include <iostream>
#include <vector>
#include <list>
using namespace std;

int ar[102][102],viz[102];
int n;
list <int> a;
int dfs(int s)
{
    viz[s] = 1;
    for (int i=1;i<=n;i++)
    {

        if (ar[i][s] && !viz[i]) {dfs(i);  }//a.push_back(i);}
    }
    return 0;
}

int main()
{
    cin >> n;
    int z;
    for (int i=1;i<=n;)
    {
        cin >> z;
        if (z == 0) i++;
        else ar[i][z] = 1;
    }
    dfs(1);
    for (int i=1;i<=n;i++)
        if(!viz[i]) a.push_back(i);
    cout << a.size() << endl;
    for (int n : a) {
        cout << n << " ";
    }
    return 0;
}


maybe i didn't understand problem properly,some help?

Here is my solution and used test cases. I don't understand why I
always get WA. 8-(

Test 1.
5 4 2 0 1 2: NO

Test 2.5 4 2 0 1 5: YES

Test 3.
5 4 2 0 3 5: YES

Test 4.
5 4 2 0 3 4: YES

Test 5.
5 5 2 0 3 4 5: YES

Test 6.
5 5 2 0 2 4 5: YES

Test 7.
5 6 2 1 2 4 5 0: NO

Test 8.
5 4 1 2 2 3: YES

Test 9.
5 5 1 2 3 4 3: YES

Test 10.
5 5 1 1 0 3 4: NO

Test 11.
5 3 0 0 1: NO

Test 12.
5 3 0 1 1: NO

Test 13.
5 3 1 2 2: YES

Test 14.
5 3 1 1 2: YES

Test 15.
5 3 4 4 5: NO

Test 16.
5 3 3 4 4: YES

Test 17.
5 3 4 5 5: NO

Test 18.
5 3 3 4 5: YES

Test 19.
2 2 1 1: YES

Test 20.
2 2 2 2: NO

Source code:

[code deleted]

Edited by moderator 13.02.2007 20:57

there are n * m possible sums. to find a solution, we can try to create such vectors such that all n * m sums are different.
Suppose the 1st vector is 1 2 3 ... n

Then the first element of next vector is taken to be n + 1(we require all unique).
Now we need to find the next element while making all n*p sums different, where p is the size of the 2nd vector.

n+1
1 2 3 ... n

initially all sums(n+2 to 2n+1) are distinct. Now we need to add new element such that
smallest possible new sum is greater than the previous sum
i.e. if new element is p
p+1>2n+1
or p >= 2n + 1
in general p+1>prev+n
p>=prev+n

what is wrong here?

for (int i =0; i< 15;++i)
{
int p = primes[i];
if ( c % p == 0)
{
   unsigned long long M = c / p;
   unsigned long long M_inv = inverse[i][ M % p ] ;
   unsigned long long h = M_inv * m[ i ];
   // r = (r + h * M ) % c

   unsigned long long high_M = (M >> 32 ) & 0xFFFFFFFFULL;
   unsigned long long low_M = ( M & 0xFFFFFFFFULL);

   // h * M= h * (high_M * 2^32 + low_M)
   unsigned long long high_h = high_M * h;
   unsigned long long e = (high_h >> 32) & 0xFFFFFFFFULL;
   unsigned long long f = (high_h & 0xFFFFFFFFULL);

   // high_h = e*2^32 + f

// h*M = h*(high_M*2^32 + low_M) = high_h * 2^32 + low_M * h =
// = (e*2^32 + f)*2^32 + low_M * h = e*2^64 + f*2^32 + low_M * h

    e = (e << 48) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 1 ) % c;

   e = (e + (f << 32) ) % c;
   e = (e + low_M * h ) % c;

   r = (r + e ) % c;
 }
}

You have to divide the participants into equal teams (rounded)
For example, for "15 10" test - (1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
Good luck :)

int n = in.nextInt();
        int l = 0;
        if (n > 24) {
            out.print("Glupenky Pierre");
        } else {
            for (int i = 1; i <= n; i++) {
                if (i / 5 == 0)
                    l = 0;
                if (i / 5 == 1)
                    l = 1;
                if (i / 5 == 2)
                    l = 8;
                if (i / 5 == 3)
                    l = 6;
                if (i / 5 == 4)
                    l = 9;

                if (i % 5 == 0)
                    l = l * 10;
                if (i % 5 == 1)
                    l = l * 10 + 1;
                if (i % 5 == 2)
                    l = l * 10 + 8;
                if (i % 5 == 3)
                    l = l * 10 + 6;
                if (i % 5 == 4)
                    l = l * 10 + 9;
                if (l/10==0)
                    out.print("0");
                out.print(l);
                out.print(" ");
            }
        }

Edited by author 07.11.2016 22:26

Edited by author 07.11.2016 22:27

There is my code:

Program t1134;

Const MaxN=1000;

Var Card      : array[1..MaxN]of boolean;
    Numb      : array[1..MaxN]of integer;
    Use       : array[1..MaxN]of boolean;
    M,N,i,j   : integer;
    ex        : boolean;

begin
Read(N,M);
if (M>N)or(m=0) then begin
 writeln('NO');
 halt(0);
end;
for i:=1 to N do Card[i]:=false;
for i:=1 to M do Use[i]:=false;
for i:=1 to M do begin
 read(Numb[i]);
 if (Numb[i]<0)or(Numb[i]>n) then begin
   writeln('NO');
   halt(0);
 end;
end;
for i:=1 to M do
 if Numb[i]=0 then begin
  Use[i]:=true;
  if Card[1] then begin
    writeln('NO');
    halt(0);
  end;
  Card[1]:=true;
 end else
 if Numb[i]=n then begin
  Use[i]:=true;
  if Card[n] then begin
    writeln('NO');
    halt(0);
  end;
  Card[n]:=true;
 end;
for i:=1 to m-1 do if not(use[i]) then
 for j:=i+1 to m do if not(use[j]) then
  if numb[i]=numb[j] then begin
   if (card[numb[i]])or(card[numb[j]-1]) then begin
    writeln('NO');
    halt(0);
   end;
   card[numb[i]-1]:=true;
   card[numb[i]]:=true;
   use[i]:=true;
   use[j]:=true;
  end;
repeat
ex:=true;
for i:=1 to m do if not(use[i]) then
 if (card[numb[i]])or(card[numb[i]-1]) then begin
  if (card[numb[i]])and(card[numb[i]-1]) then begin
    writeln('NO');
    halt(0);
  end;
  if card[numb[i]] then begin
   card[numb[i]-1]:=true;
   use[i]:=true;
  end else begin
   card[numb[i]-1]:=true;
   use[i]:=true;
  end;
 end;
until ex;
Writeln('YES');
end.

does anyone knows test 6?

http://pastebin.com/af0DFBg0
Look at my code please.
If someone could say me what wrong, it would be great.
Thank you!

У задачи отклеилось описание на русском? Хоть и шарю по английскому, всё равно неприятно тратить несколько минут на перевод...

Edited by author 20.02.2016 16:49

i see N%3 solution everywhere but does anyone have any explanation as to why the guy who has mod-3 no of stones should lose. Some logical or mathematical reason would be much appreciated.