Wat is the difference between primary storage, secondary storage, and off-line storage
1 answer:
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Answer:
Here is the Python code:
runs=int(input("how many runs do you want to add items to dictionary? ")) #store input number of runs
d = dict() #creates a dictionary
d = {0:10,1:20} #old dictionary
count=0 #counts number of runs
for x in range(2,runs+2): #loop to append key/value pairs to dictionary
d[x]=(x*10)+10 #multiples and adds 10 to the each value
count+=1 #adds 1 to the count at each iteration
print("After the #",count, "run the new dictionary is: ",d) #prints the new dictionary
Explanation:
I will explain the program with an example:
The old dictionary is :
d = {0:10,1:20}
runs = 3
count = 0
At first iteration:
x = 2
d[x]=(x*10)+10
This becomes:
d[2]=(2*10)+10
d[2]= 20 + 10
d[2]= 30
count+=1
count = 1
print("After the #",count, "run the new dictionary is: ",d)
This statement displays the first iteration result :
After the # 1 run the new dictionary is: {0: 10, 1: 20, 2: 30}
At second iteration:
x = 3
d[3]=(3*10)+10
This becomes:
d[3]=(3*10)+10
d[3]= 30 + 10
d[3]= 40
count+=1
count = 2
print("After the #",count, "run the new dictionary is: ",d)
This statement displays the first iteration result :
After the # 2 run the new dictionary is: {0: 10, 1: 20, 2: 30, 3: 40}
At third iteration:
x = 4
d[4]=(4*10)+10
This becomes:
d[4]=(4*10)+10
d[4]= 40 + 10
d[4]= 40
count+=1
count = 3
print("After the #",count, "run the new dictionary is: ",d)
This statement displays the first iteration result :
After the # 3 run the new dictionary is: {0: 10, 1: 20, 2: 30, 3: 40, 4: 50}
Now the loop breaks as x = 5 necause n+2 = 3+2 = 5 limit is reached
The screenshot of program along with its output is attached.
