Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
<h2>Alignment section helps to wrap text in Excel</h2>
Explanation:
- Alignment section available in Home ribbon has many option such as aligning the text to left, right, center.
- It also has options to rotate the selected text diagonally, vertically, etc.
- The user can also merge two or more cells and do necessary alignment
- When it comes to wrapping the text, texts are wrapped when the count of letters exceeds the width of the cell. Excel has "Wrap text" option to wrap the word to the next line.
Answer:
count_p = 0
count_n = 0
total = 0
while True:
number = int(input("Enter an integer, the input ends if it is 0: "))
if number == 0:
break
else:
total += number
if number > 0:
count_p += 1
elif number < 0:
count_n += 1
print("The number of positives is: " + str(count_p))
print("The number of negatives is: " + str(count_n))
print("The total is: " + str(total))
print("The average is: " + str(total / (count_p + count_n)))
Explanation:
Initialize the variables, count_p represens the number of positives, count_n represents the number of negatives, and total represents the total of the numbers
Create a while loop iterates until the user enters 0. If the number is not 0, then add it to the total. If the number is greater than 0, increase count_p by 1. If the number is smaller than 0, increase count_n by 1.
When the loop is done, print the count_p, count_n, total, and average
