I used to use pixlr all the time so i should know.
Ok so
1 is picker
2 is lasso select
3 is crop
4 is Liquefy
5 is clone
6 is dodge/burn
7 is Pen
8 is Eraser
9 is Gradient
10 is Text
11 is Zoom
12 is Switch Colors
13 is Marquee Select
14 is Wand select
15 is Cutout/Mask
16 is Heal
17 is Blur/Sharpen
18 is Sponge Color
19 is draw
20 is Fill
21 is I don’t know I have never seen that before
22 is Picker
So sorry if it took a while i did not know what the star was
In an if...else statement, if the code in the parenthesis of the if statement is true, the code inside its brackets is executed. But if the statement inside the parenthesis is false, all the code within the else statement's brackets is executed instead.
Of course, the example above isn't very useful in this case because true always evaluates to true. Here's another that's a bit more practical:
#include <stdio.h>
int main(void) {
int n = 2;
if(n == 3) { // comparing n with 3 printf("Statement is True!\n");
}
else { // if the first condition is not true, come to this block of code
printf("Statement is False!\n"); } return 0;
}
Output:
Statement is False!
B) Raw materials; As stated, European powers did indeed grow more industrialized, therefore they needed more raw materials. The North provided lumber, spent time whaling, etc.. The South spent time growing cash crops such as rice, cotton, and indigo.
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
Answer:
There are two error in this program--
- In header file inclusion, file is not defined.
- In the statement "result = ++(num1 + num2);" , bracket is fixed after the increment operator.
Explanation:
- For the first error, the user needs to add the file because "#include" is used to add the library for the program which states about the function and symbols used in the program.
- The second error is because there must be a variable with the increment operator ( increment operator is being used to increase the value of a variable by 1), but there is a small brace fix in between the operator and operands.
