mirror of
https://github.com/Xevion/contest.git
synced 2025-12-11 08:06:53 -06:00
questions 8, 9
This commit is contained in:
@@ -4,7 +4,7 @@
|
|||||||
|
|
||||||
This can be guessed on pretty easily by analyzing the magnitudes of the powers. Every number here is within a mildly close range of eachother, and the highest numbers have some of the *lowest* powers, so we can assume that the one with the highest power will be the largest when evaluated.
|
This can be guessed on pretty easily by analyzing the magnitudes of the powers. Every number here is within a mildly close range of eachother, and the highest numbers have some of the *lowest* powers, so we can assume that the one with the highest power will be the largest when evaluated.
|
||||||
|
|
||||||
# Question 2
|
## Question 2
|
||||||
|
|
||||||
This is simple String and Integer addition.
|
This is simple String and Integer addition.
|
||||||
|
|
||||||
@@ -88,13 +88,31 @@ It's pretty easy modulus, I got `ABE`.
|
|||||||
|
|
||||||
## Question 9
|
## Question 9
|
||||||
|
|
||||||
Just a little
|
Solve from left to right, (P)(E)(MD)(AS).
|
||||||
|
|
||||||
|
Tip: You can solve in sections with the parentheses above, as multiplication/division can be done in any order together without changing the result. It helps a little bit with cutting down the manual solving time, if you ask me.
|
||||||
|
|
||||||
```java
|
```java
|
||||||
|
3 + 5 / 2 + 2.0
|
||||||
|
3 + 2 + 2.0
|
||||||
|
5 + 2.0
|
||||||
|
7.0
|
||||||
```
|
```
|
||||||
|
|
||||||
## Question 10
|
## Question 10
|
||||||
|
|
||||||
|
Just a little bit of math and tracking of a changing array.
|
||||||
|
|
||||||
|
```java
|
||||||
|
r[3] = 19;
|
||||||
|
r[1] = r[3] * 2 = 38;
|
||||||
|
r[4] = r[1] / 2 = 19;
|
||||||
|
r[2] = r[4 % 3 = 1] / 3 = 38 / 3 = 12;
|
||||||
|
```
|
||||||
|
```java
|
||||||
|
r = [0, 38, 12, 19, 19]
|
||||||
|
```
|
||||||
|
|
||||||
## Question 11
|
## Question 11
|
||||||
## Question 12
|
## Question 12
|
||||||
## Question 13
|
## Question 13
|
||||||
|
|||||||
Reference in New Issue
Block a user