Participate in the Free Grand Raffle draw and win Uni-currency. The value of a university currency equals one US dollar on the day of the raffle.

PARTICIPATE IN THE FREE GRAND RAFFLE DRAW AND WIN UNI CURRENCY. THE VALUE OF A UNI CURRENCY WOULD BE EQUAL TO A US DOLLAR ON THE DAY OF THE RAFFLE DRAW.

The Unigroup wants its single currency (the digital currency) to go into the hands of so many people around the world that it will free the single currency to all around the world through a unique system that takes place through participation in a raffle distributed

Thousands of people are already from around the world. Remember that you have lost nothing, because it is free. So do not miss this free opportunity, but quickly enter the raffle of the big raffle and get a 100% chance to win the university currencies.

Click on the link – Uni-currency

dom – XSS with escape equals sign in the jQuery selector

The site uses jQuery 1.8.3, which has a known XSS vulnerability in the selector. (Https://snyk.io/vuln/npm:jquery:20120206).

It happens filtered and urdecoded ducument.location.hash (val2 below) value within the voter.

$ (& Div; div[data-foo=''+filter(val1)+''][data-value=''+filter(val2)+'']& # 39)

Function filter (str) {
if (str)
return str.replace (/ ([ #;?%&,.+*~':"!^$[]() => |  / @]) / g, & # 39; \ $ 1 & # 39;
return str;
}

I have reached the following payload:


It would work if the equals sign were not replaced by =, Browsers do not seem to tolerate = at all.

Any ideas how this can be bypassed? Or maybe another payload would work here?

Why do I get "java.lang.NullPointerException" each time I use equals ()? Java Poo

Well, I'm doing Poo programs and every time I want to use the equals () method, when I run the code, I jump red: "java.lang.NullPointerException", why is it going to be like this?

public class user {

private string user;
private string password;

public user (string name, string password) {

this.username = name;
this.password = password;
}

public String getName () {

return this.user;
}

public void setName (stringname) {

this.username = name;
}


public String getPassword () {


Please return this password.
}

public void setPassword (String password) {


this.password = password;
}   

}

public class sistema {

private string name;
private user list[];
int usersadded = 0;

public system (string name, int number of users) {

this.name = name;
this.listDeUsers = new user[cantidadDeUsuarios];

}

public String getName () {

return this name;

}

public void setName (stringname) {

this.name = name;
}

public user[] getListaDeAutos (user list of users) {

returns this list of users;
}

public void setUserList (User[] User list) {

this.list of users = list of users;

}

public boolean loguearUser (string user, string password) {

Boolean state = false;
int i = 0;

for (i = 0; i <userlist.length; i ++) {
if (list of users[i].getName (). equals (user)
&& user list[i].getPassword (). equals (password)) {

State = true;
}

miscellaneous

State = wrong;
}
}

Returning state;
}

public void addUsers (user user) {

Member list[usuariosAgregados] = User;
User aggregates ++;

}

}

Python – For certain numbers, find triplets where the product of two numbers equals the third number

I am trying to solve a problem of the coding challenge, and the question is described as follows

With N integers A1, A2, …, AN, count the number of triplets (x, y,
z) (with 1≤x <y <z≤N), so that at least one of the following applies
true:

Ax = Ay × Az and / or

Ay = Ax × Az and / or

Az = Ax × Ay

5 2 4 6 3 1

In example case # 1, the only triplet satisfying the condition stated in the problem statement is (2, 4, 5). The triplet is valid because the second, fourth and fifth integers are 2, 6 and 3 and 2 x 3 = 6. So the answer here is 1

2 4 8 16 32 64

The six triplets fulfilling the condition stated in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). The answer here is 6

My code in Python:

import itertools
count = 0
for t in itertools.combinations (l, 3):
if t[0]* t[1]== t[2] or t[1]* t[2]== t[0] or t[0]* t[2]== t[1]:
count + = 1
print (count)

This is the naive way to create all possible 3 length combinations and to check the condition. This works well for smaller inputs, but increases with the complexity of the inout size. I go for an example of this 1,2,3,6,8 The generated combinations are (2,3,6), (2,3,8) 2,3,6 meet the condition, so that the check on 2,3,8 is not required and can be avoided. How can I change my code to use this observation?

Field Theory – How to prove that $ left {t ^ 2, t ^ 3 right } $ equals the vanishing amount of $ y ^ 2-x ^ 3 $?

Exercise 3.2 of Hartshorne is about proving that morphisms of varieties of homeomorphisms can be subordinated without being isomorphisms of varieties. The considered morphism is $ varphi: t mapsto (t ^ 2, t ^ 3) $ where the picture is intended as a curve $ y ^ 2-x ^ 3 $,

I do not understand what fields $ Bbbk $ the sentence $ left {t ^ 2, t ^ 3 right } $ equal the vanishing set of $ y ^ 2-x ^ 3 in Bbbk[x,y]$ – Only why it is included in the disappearing set. If so $ (a, b) $ satisfied $ a ^ 3 = b ^ 2 $ We need at least the existence of square / cubic roots, as we want $ t $ so that $ t ^ 2 = a, t ^ 3 = b $, Suppose the field is algebraically closed. Then we have a creature worthy of the name $ sqrt a $ that satisfies $ sqrt a ^ 2 = a $ and under the assumption $ sqrt a ^ 6 = b ^ 2 $, But why should we do that? $ sqrt a ^ 3 = b $? (more precisely, why can we choose $ sqrt a $ have this property?)

Maybe this is a basic field / Galois theory, but better late than never.