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eimsori
1 month ago
4

Given an integer representing a 10-digit phone number, output the area code, prefix, and line number using the format (800) 555-

1212. Ex: If the input is: 8005551212 the output is: (800) 555-1212 Hint: Use % to get the desired rightmost digits. Ex: The rightmost 2 digits of 572 is gotten by 572 % 100, which is 72. Hint: Use // to shift right by the desired amount. Ex: Shifting 572 right by 2 digits is done by 572 // 100, which yields 5. (Recall integer division discards the fraction). For simplicity, assume any part starts with a non-zero digit. So 0119998888 is not allowed.
Computers and Technology
2 answers:
amid [910]1 month ago
5 0

When provided with an integer for a 10-digit phone number, format the output to display the area code, prefix, and line number as (800) 555-1212.

Examples have been considered in the code preparation.

The following code corresponds to the statement above:

Details:

phone_number = input()

if (len(phone_number)==10):

print(phone_number[0] + phone_number[1] + phone_number[2] + '-' + phone_number[3] + phone_number[4] + phone_number[ 5] + '-' + phone_number[6] + phone_number[7] + phone_number[8] + phone_number[9])

else:

print('invalid phone number')

An integer is inputted, and its length is validated.

A number that begins with 0 is not permissible.

amid [910]1 month ago
5 0

Response:

input: 8005551212

phone_number = int(input())

area_code = phone_number // 10000000

prefix = (phone_number // 10000) % 1000

line_number = phone_number % 10000

print(f'({area_code}) {prefix}-{line_number}')

Details:

#Input data

8005551212

#Assign phone number to input data

phone_number = int(input())

#Logic for determining area_code, prefix, and line_number

area_code = phone_number // 10000000

prefix = (phone_number // 10000) % 1000

line_number = phone_number % 10000

#f-strings(format strings) provide a more concise alternative for str.format() calls.

print(f'({area_code}) {prefix}-{line_number}')

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The provided question is lacking details. It can be retrieved from search engines. However, please see the full question below:

Question

It points out a mistake in using a part of the performance equation as a measure of performance. For example, examine these two processors. P1 operates at a clock frequency of 4 GHz, has an average CPI of 0.9, and needs to execute 5.0E9 instructions. P2 runs at 3 GHz, with an average CPI of 0.75, needing to execute 1.0E9 instructions. 1. A common misunderstanding is assuming that the processor with the highest clock rate has the best performance. Determine if this holds true for P1 and P2. 2. Another misconception is that the processor with the greater number of executed instructions necessarily has a longer CPU time. If processor P1 processes 1.0E9 instructions and both processors have unchanged CPI values, calculate how many instructions P2 can complete in the same duration that P1 uses to execute 1.0E9 instructions. 3. A frequent error is to use MIPS (millions of instructions per second) to evaluate the performance of different processors, believing that the one with the highest MIPS is the best. Verify whether this applies to P1 and P2. 4. MFLOPS (millions of floating-point operations per second) is another common metric, defined as MFLOPS = No. FP operations / (execution time x 1E6), but it suffers from the same issues as MIPS. Assuming 40% of the instructions executed on both P1 and P2 are floating-point instructions, calculate the MFLOPS values for the programs.

Answer:

(1) We apply the following formula:

                                         CPU time = number of instructions x CPI / Clock rate

By substituting 1 GHz = 10⁹ Hz, we find:

CPU time₁ = 5 x 10⁹ x 0.9 / 4 GHz

              = 4.5 x 10⁹ / 4 x 10⁹ Hz = 1.125 s

and,

CPU time₂ = 1 x 10⁹ x 0.75 / 3 GHz

                    = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

This shows that P2 is significantly faster than P1 because CPU₂ is shorter than CPU₁

(2)

Determine the CPU time of P1 using (*)

CPU time₁ = 1 x 10⁹ x 0.9 / 4 GHz

                  = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

Next, we need the count of instructions₂ so that CPU time₂ = 0.225 s, applying (*) with clock rate₂ = 3 GHz and CPI₂ = 0.75

Thus, instruction count₂ x 0.75 / 3 GHz = 0.225 s

Consequently, instruction count₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸

Thus, P1 can handle more instructions than P2 within the same time frame.

(3)

We remember that:

MIPS = Clock rate / CPI x 10⁶

 So, MIPS₁ = 4 GHz / 0.9 x 10⁶ = 4 x 10⁹ Hz / 0.9 x 10⁶ = 4444

MIPS₂ = 3 GHz / 0.75 x 10⁶ = 3 x 10⁹ / 0.75 x 10⁶ = 4000

This indicates that P1 has a higher MIPS

(4)

 Now we note that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Accordingly,

                    MFLOPS₁ = 1777.6

                    MFLOPS₂ = 1600

Again, P1 boasts a greater MFLOPS

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