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miv72 [106K]
3 years ago
12

Hamilton is the best

Computers and Technology
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

Yes he is

Explanation:

You might be interested in
Write a Python program that takes a file name as input and generates the following output: File size in bytes and KBs Time the f
Airida [17]

Answer:

See explaination

Explanation:

import os

import time

from stat import *

file = input("Enter file name: ")

details = os.stat(file)#stores statistics about a file in a directory

print("File size: ", details[ST_SIZE])#retreiving size

print("File last accessed time: ", time.asctime(time.localtime(details[ST_ATIME])))#access time

print("File last modified time: ", time.asctime(time.localtime(details[ST_MTIME])))#modified time

print("File creation time: ", time.asctime(time.localtime(details[ST_CTIME])))#creation time

print("First character: ", file[0])

print("Middle character: ", file[len(file) // 2])

print("Last character: ", file[len(file) - 1])

num_words = 0

num_lines = 0

f=open(file, "r")

lines = f.readlines()

for line in lines:

num_lines = num_lines + 1

num_words = num_words + len(line.split())#add no. of words in that line

print("Number of lines: ", num_lines)

print("Number of words: ", num_words)

8 0
3 years ago
1. Scrieţi un program care citeşte un număr natural n şi determină produsul cifrelor impare ale lui n. De exemplu, pentru n = 23
saveliy_v [14]

Answer:

1.  

num1 = input("Enter the value of n")  

n = int(num1)  

def calc(n):  

Lst = []  

while n > 0:  

remainder = n % 10  

Lst.append(remainder)  

quotient = int(n / 10)  

n = quotient  

i = len(Lst) - 1  

sum = 1  

for i in range(0, len(Lst)):  

if(i % 2 != 0):  

sum *=Lst[i];  

else:  

continue  

print(sum)  

return(0)  

\ r2

num=input("Enter the value of n")

arr=[12,-12,13,15,-34,-35,35,42]

def div7positive():

  sum = 0

  m = 0

  i = 0

  while m <= 7:

      if(arr[i]>=0 and arr[i]%7 == 0):

          sum +=arr[i]

      i = i + 1

      m += 1

       

   

  print("sum of number divisible by 7 and positive is", +sum);

div7positive()

\ r

\ r3.Write an algorithm that reads a natural number n and calculates the sum:

\ rS = 1 / (1 * 2) + 1 / (2 * 3) + 1 / (3 * 4) +… + 1 / ((n-1) * n)

\ r

num=input("Enter the value of n")

def sumseries():

  sum = 0

  m= 0

  while m <= 7:

      sum +=1/((int(num)-1)*int(num))

      m += 1

  print("sum of series is", +sum)

sumseries()

\ R4. Read a natural number n. Calculate the sum of its own divisors n. For example, for n = 12, the sum of its own divisors is 2 + 3 + 4 + 6 = 15

num=input("Enter the value of n")

def sumdivisors():

  sum = 0

  m= 2

  while m <= int(num):

      if int(num) % m == 0:

          sum += m

      m += 1

  print("sum of divisors is", +sum)

sumdivisors()\ r

\ R5. We read a natural number n and then whole numbers. Calculate and display the sum of the natural numbers between 10 and 100. For example, if n = 5 and then read 30, –2, 14, 200, 122, then the sum will be 44 (that is, 30 + 14).

\ r

Lst = []  

num=input("Enter the value of n")

i = 0

while i<= int(num):

  num1=input("Enter the element of array")

  Lst.append(int(num1));

  i += 1

def numbet0and100():

  sum = 0

  m= 0

  while m <= int(num):

      if Lst[m] <= 100 or Lst[m] >=0:

          sum += Lst[m]

      m += 1

  print("sum of numbers between 0 and 100 is", +sum)

numbet0and100()

\ R6. A natural number n of maximum 4 digits is read. How many digits are in all numbers from 1 to n? For example, for n = 14 there are 19 digits, and for n = 9 there are 9 digits.

\ r

num = input("Enter the value of n")

n = int(num)

print(“sum as required is:” (n -9) + n)  

 

\ R7. Read the natural numbers n and S, where n can be 2, 3, 4 or 5. Show all the numbers of n digits that have the numbers in strictly ascending order, and the sum of the digits is S. For example, for n = 2 and S = 10, 19, 28, 37, 46 will be displayed.

num = input("Enter the value of n")

n = int(num)

Lst = []

def calc(n):

  i = 1

  total = pow(10,n)

  while i <= total:

      j = i

      while j > 0:

          remainder = j % 10  

          Lst.append(remainder)

          quotient = int(j / 10)

          if quotient > 0:

              j = quotient

          else:

              length = len(Lst) - 1

              sum = 0

              while length >= 0:

                  sum += Lst[length]

                  length = length - 1

              if sum == pow(10,n):

                  print(j)

              k = len(Lst)

              del Lst[0:k]

      i = i + 1

  return(0)

calc(n)

\ r

\ R8. We consider the row 1, 1, 2, 3, 5, 8, 13, ... in which the first two terms are 1, and any other term is obtained from the sum of the preceding two. Write an algorithm that reads a natural number n and displays the first n terms of this string. For example, for n = 6, 1, 1, 2, 3, 5, 8 will be displayed.

\ r

nterm = int(input("What number of terms do you want?"))

a, b = 0, 1

totalcount = 0

if nterm <= 0:

 print("Please input a positive number")

elif nterm == 1:

 print("Fibonacci number upto which",nterm,":")

 print(a)

else:

 print("Fibonacci series:")

 print(nterm)

 while totalcount < nterm:

     print(a)

     nth = a + b

     a = b

     b = nth

     totalcount += 1

 

\ 9. Write an algorithm that reads two natural numbers n1 and n2 and displays the message "yes" if the sum of the squares of the digits of n1 is equal to the sum of the numbers of n2 or "no" otherwise. For example, for n1 = 232 and n2 = 881, "yes" will be displayed, and for n1 = 45 and n2 = 12, "no" will be displayed.

num1 = input("Enter the value of n")

num3 = input("Enter the value of n")

n = int(num1)

num2 = int(num3)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      quotient = int(n / 10)

      n = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum += Lst[i]* Lst[i]

      i = i - 1

  return(sum)

def calc1(num2):

  Lst = []

  while num2 > 0:

      remainder = num2 % 10  

      Lst.append(remainder)

      quotient = int(num2 / 10)

      num2 = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum = Lst[i] + Lst[i]

      i = i - 1

  return(sum)

a = calc(n)

b = calc1(num2)

if a == b:

  print("yes")

else:

  print("No")

\ r

\ R10. Write an algorithm that reads a natural number n and displays the message "yes" if all of its n numbers are distinct, or "no" if n does not have all the distinct digits. For example, for n = 37645 it will display "yes" and for 23414 it will show "no".

\ r

num1 = input("Enter the value of n")

n = int(num1)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      print( remainder)

      quotient = int(n / 10)

      n = quotient

       

  flag = 1

  for i in range(0, len(Lst)):

      for j in range(i+1, len(Lst)):    

          if Lst[i] == Lst[j]:    

              flag = 0

              break

           

  if flag == 1:

      print("YES")

  else:

      print("NO")

  return(0)

Explanation:

Please check answer.  

3 0
3 years ago
Software built and delivered in pieces is known as
mafiozo [28]

Answer:

Incremental technique is the way in which software is built and delivered in pieces. The concept is to keep the client and developer on same page and the client is known as a non tech person, so he should be given software in piece by piece to avoid any confusion and sudden change.

Agile method is the best example of this technique in which steps are defined on contract basis and the software is delivered and build by pieces to keep client and developer on same page.

8 0
3 years ago
Read 2 more answers
What is a boot sector
Neporo4naja [7]

Explanation:

A boot sector is the sector of a persistent data storage device which contains machine code to be loaded into random-access memory and then executed by a computer system's built-in firmware. Usually, the very first sector of the hard disk is the boot sector, regardless of sector size and partitioning flavor.

4 0
2 years ago
Sequential and direct access are two methods to locate data in memory. Discuss why major devices now a days use direct access? H
djverab [1.8K]

Answer:

Direct data access reduces the speed of retrieving data from memory or storage. Retrieving data and storing it in a cache memory provides direct access to data in the storage.

Explanation:

Sequential memory access, as the name implies, goes through the memory length location in search of the specified data. Direct memory access, provides a memory location index for direct retrieval of data.

Examples of direct and sequential memory access are RAM and tapes respectively. Data in sequential memory access can be access directly by getting data in advance and storing them in cache memory for direct access by the processor.

3 0
3 years ago
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