Cheap or expensive brushes / Art suppliers

Cheap or expensive brushes / Art suppliers

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  1. Cheap or expensive brushes / Art suppliers

    In quality and price, the best watercolor brushes can vary from those in a sealed package of perhaps five, hanging at the end of the aisle of your national chain hobby shop to those that have a real stamp of approval as "By command, Her Majesty the Queen." (Famously, Queen Victoria ordered Winsor & Newton what would be her 7 favorite size in the series 7 of the red saber rounds Kolinsky). The first option could serve, for a short time, to clean a child's boots in his wardrobe. The latter could be framed with pride on the wall of your studio. In between, you will find a wide selection of more than just repairable brushes, of recognized fine brands. Some of these brands are: Winsor & Newton Series 7, Isabey, Rafael, Arches, Escoda, Pro Arte and others. The smaller lines, but still quite useful, include: Winsor & Newton Series 666, Princeton and Grumbacher. I recommend buying a good brand. Avoid the various gangue packages; they will perform so poorly in practice and in durability that they will feel very discouraged. I have some of each mentioned brand and I enjoy every brush. And yes, I bought cheappo brushes in the past … and I had to throw them away.

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ray network – Who pays the fees in the chain per settlement transaction?

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Problem with String in C, exclusion of term in the chain, through string comparison

I elaborated a code that excluded the similar term in the two chains, but it does not print the result of the text by removing the second term, it only removes the first term of the first string.

"Create a program that receives a phrase and a term.
the term is present in the phrase, remove it and print the phrase
modified, otherwise, display "Term not found in the
phrase ". Consider the maximum chain size equal to 200."

# include 
# include 

int main () {} ()
char text[200];
char word[10];
In this case,
j = 0, i, c = 0, R;
scanf ("% s", text);
gets (text);
scanf (word "% s");
Tam = strlen (string);
for (i = 0; i <L1; i ++) {
r = strcmp (text, word);
j = j + 1;
if ((text! = & # 39;  0 & # 39;) && (word! = & # 39;  0 & # 39;)) {
for (i = 0; i <L1; i ++) {
ind = (strlen (string) -1) / j;
j = j + 1;
printf ("% s  n", text);

} else
printf ("Term not found in the phrase.  n");
return 0;

c – How do I compare individual elements in a chain?

"I do not know how to write the loop to compare 100 possible colors and have separate counters for each one."

You can have separate counters for each one by defining a matrix for the counters. Conveniently for you, "colors" are integers, so your matrix index can serve as the "color" you are counting. The stored value (in each element of the matrix) will be the number of times you have seen that "color" in the entry.

Then, go through the entry and increase the corresponding element in your counters matrix. In this way, it does not matter if the match of a given sock is the next in line in the input or several input elements away, and you will not have to do anything costly, like comparing each input element with any other input element.

At the end of that loop, repeat over your counter array; the value in each item will tell you about that "color" sock:

0: There are no socks of that color in the entrance.
odd number: there are `x / 2` pairs of this color and an odd sock
even number: there are `x / 2` pairs of this color

Count the number of pairs and odd socks as you go, and at the end you have the total number of pairs of odd socks and socks, which I guess is the way out you want.

javascript – How to chain XSS loads?

I found a parameter on a website that is vulnerable to XSS. The website is a game and the parameter in question is the name of a player's team. However, the parameter is limited to only 20 characters. There is another page on the website where these teams of players can be displayed. I can make several teams of players and show their names on this page.

I was able to create an alert box using the following team names on three separate computers:

The comments omit any HTML between the names of the teams, therefore, they "chain" and show the alert box. However, nothing really malicious can be done with this.

I wanted to implement the theft of cookies, so I first created this payload (the IP address has been changed, except the value "169", and the number of characters of the IP address is identical to the actual IP address):

Of course, this would not fit in 20 characters, so I divided it that way (I can create up to 8 teams of players to show on the page):

(the words "document" and "69" should be separated since the site's filter considers them inappropriate)

However, I encountered some problems in doing this. First, the bars in http: // They are interpreted as comments and begin to comment on unwanted things. Second, whenever there is a & # 39; symbol, any comment after it is ignored. For example: Comment does not work

How can I avoid this and join the XSS chain?

c – Function to invert a chain

I tried to make a function to invert a string but the program always returns

char reverse (char string)[])
int length, i;

length = strlen (string);

char reverse_string[length];

(i = 0; i <length; i ++)
reverse_string[i] = string[(length - i) - 1];

return reverse_string;


int main ()
char string[50];

gets (string);

printf ("% s", reverse (string));

return 0;


I thought I could be placing the character & # 39; 0 & # 39; in the first position of reverse_string, but I added the "-1" in the operation and even then the program always returns (null). If anyone knows what is happening and can give advice, I appreciate it!

Where is the code of reorganization of the chain in Bitcoin Core?

I am trying to understand how the Bitcoin kernel works in depth. Can anyone tell me where the code is executed during a blockchain reorganization in Bitcoin Core?

Any explanation would also be greatly appreciated.

How to host a website in the Bitcoin or Bitcoin SV block chain?

I saw somewhere that there is a way to host a website in a bitcoin SV transaction. How can I do it? I want to do it without 3d party service. Is there any guidance on how to place an html page in a bitcoin SV transaction?

performance – Python program to solve the multiplication of the matrix chain

A homework assignment in the school required that I write a program for this task:

In the problem of multiplication of the matrix chain, we are given a sequence of
matrices A (1), A (2), ..., A (n). The objective is to calculate the product.
A (1) ..., A (n) with the minimum number of scalar multiplications.
Therefore, we have to find an optimal parenthesis of the matrix.
product A (1) ..., A (n) such that the cost of computing the product is

Here is my solution for this task (in Python):

def matrix_product (p):
Return m and s.

subway[i][j]    is the minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
length = len (p) # len (p) = number of matrices + 1

# m[i][j]    is the minimum number of multiplications needed to calculate the
# product of matrices A (i), A (i + 1), ..., A (j)
# s[i][j]    It is the matrix after which the product is divided into the minimum.
# number of multiplications needed
m = [[-1]* length for _ in range (length)]s = [[-1]* length for _ in range (length)]matrix_product_helper (p, 1, length - 1, m, s)

return m, s

def matrix_product_helper (p, start, end, m, s):

Return the minimum number of scalar multiplications needed to calculate the
product of matrices A (start), A (start + 1), ..., A (end).

The minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j) is stored in m[i][j].

The index of the matrix after which the previous product is divided into an optimal
the parenthesis is stored in s[i][j].

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
yes m[start][end]    > = 0:
return m[start][end]




    if start == end:
q = 0
q = float (& # 39; inf & # 39;)
for k in range (start, end):
temp = matrix_product_helper (p, start, k, m, s) 
+ matrix_product_helper (p, k + 1, final, m, s) 
+ p[start - 1]*P[k]*P[end]
            if q> temp:
q = temperature
s[start][end]    = k

subway[start][end]    = q
come back

def print_parenthesization (s, start, end):
Print the optimal parenthesis of the matrix product A (start) x
A (start + 1) x ... x A (end).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.
if start == end:
print (& # 39; A[{}]& # 39; .format (start), end = & # 39; & # 39;)
he came back

k = s[start][end]




    print (& # 39 ;, & end = & # 39; & # 39;)
print_parenthesization (s, start, k)
print_parenthesization (s, k + 1, final)
print (& # 39;) & # 39 ;, end = & # 39; & # 39;)

n = int (entry (& # 39; Enter the number of arrays: & # 39;))
p = []
for i in the range (n):
temp = int (entry (& # 39; Enter the number of rows in the array {}: & # 39; .format (i + 1)))
p.append (temp)
temp = int (entry (& # 39; Enter the number of columns in the array {}: & # 39; .format (n)))
p.append (temp)

m, s = matrix_product (p)
print (& # 39; The number of scalar multiplications needed: & # 39 ;, m[1][n])
print (& # 39; Optimum parenthesis: & # 39 ;, end = & # 39; & # 39;)
print_parenthesization (s, 1, n)

Here is an example of output:

Enter the number of matrices: 3
Enter the number of rows in the matrix 1: 10
Enter the number of rows in the matrix 2: 100
Enter the number of rows in the matrix 3: 5
Enter the number of columns in the matrix 3: 50
The number of scalar multiplications needed: 7500.
Optimum parenthesis: ((A[1]A[2])A[3])

NOTE – The time needed to print_parenthesization () (for this example) it is 0: 00: 15.332220 seconds.

Therefore, I would like to know if I could make this program shorter and more efficient.

Any help would be greatly appreciated.

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