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Mariana [72]
2 years ago
11

1. Scrieţi un program care citeşte un număr natural n şi determină produsul cifrelor impare ale lui n. De exemplu, pentru n = 23

453, produsul este 45.
2.Scrieţi un algoritm care citeşte un număr natural n apoi n numere întregi. Să se determine câte din cele n numere întregi sunt strict pozitive şi divizibile cu 7. De exemplu, pentru n = 7 şi numerele -28, 21, 77, 15, -1, 42, -7 se va afişa 3 (acestea sunt 21, 77, 42).

3.Scrieţi un algoritm care citeşte un număr natural n şi calculează suma:
S = 1/(1*2) + 1/(2*3) + 1/(3*4) + … + 1/((n-1)*n)

4. Se citeşte un număr natural n. Să se calculeze suma divizorilor proprii ai lui n. De exemplu, pentru n=12, suma divizorilor proprii este 2+3+4+6=15

5. Se citeşte un număr natural n şi apoi n numere întregi. Să se calculeze şi să se afişeze suma numerelor naturale cuprinse între 10 şi 100. De exemplu, dacă n=5 şi apoi se citesc 30, –2, 14, 200, 122, atunci suma va fi 44 (adică 30 + 14).

6. Se citeşte un număr natural n de maximum 4 cifre. Câte cifre sunt în toate numerele de la 1 la n? De exemplu, pentru n=14 există 19 cifre, iar pentru n=9 sunt 9 cifre.

7. Se citesc numerele naturale n şi S, unde n poate fi 2, 3, 4 sau 5. Să se afişeze toate numerele de n cifre care au cifrele în ordine strict crescătoare, iar suma cifrelor este S. De exemplu, pentru n=2 şi S=10, se vor afişa 19, 28, 37, 46.

8. Se consideră şirul 1, 1, 2, 3, 5, 8, 13, ... în care primii doi termeni sunt 1, iar orice alt termen este obţinut din suma precedenţilor doi. Scrieţi un algoritm care citeşte un număr natural n şi afişează primii n termeni ai acestui şir. De exemplu, pentru n = 6, se vor afişa: 1, 1, 2, 3, 5, 8.

9. Să se scrie un algoritm care citeşte două numere naturale n1 şi n2 şi care afişează mesajul „da” dacă suma pătratelor cifrelor lui n1 este egală cu suma cifrelor lui n2 sau „nu” în caz contrar. De exemplu, pentru n1=232 şi n2=881 se va afişa „da”, iar pentru n1=45 şi n2=12 se va afişa „nu”.

10. Să se scrie un algoritm care citeşte un număr natural n şi afişează mesajul „da” dacă toate cifrele lui n sunt distincte, sau „nu” dacă n nu are toate cifrele distincte. De exemplu, pentru n=37645 se va afişa „da”, iar pentru 23414 se va afişa „nu”.
Computers and Technology
1 answer:
saveliy_v [14]2 years ago
3 0

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.  

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Write a Java class to perform the following: 1. Write a method to search the following array using a linear search, ( target ele
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Answer:

Check the explanation

Explanation:

Linear search in JAVA:-

import java.util.Scanner;

class linearsearch

{

  public static void main(String args[])

  {

     int count, number, item, arr[];

     

     Scanner console = new Scanner(System.in);

     System.out.println("Enter numbers:");

     number = console.nextInt();

   

     arr = new int[number];

     System.out.println("Enter " + number + " ");

     

     for (count = 0; count < number; count++)

       arr[count] = console.nextInt();

     System.out.println("Enter search value:");

     item = console.nextInt();

     for (count = 0; count < number; count++)

     {

        if (arr[count] == item)

        {

          System.out.println(item+" present at "+(count+1));

         

          break;

        }

     }

     if (count == number)

       System.out.println(item + " doesn't found in array.");

  }

}

Kindly check the first attached image below for the code output.

Selection Sort in JAVA:-

public class selectionsort {

   public static void selectionsort(int[] array){

       for (int i = 0; i < array.length - 1; i++)

       {

           int ind = i;

           for (int j = i + 1; j < array.length; j++){

               if (array[j] < array[ind]){

                   ind = j;

               }

           }

           int smaller_number = array[ind];  

           array[ind] = array[i];

           array[i] = smaller_number;

       }

   }

     

   public static void main(String a[]){

       int[] arr = {9,94,4,2,43,18,32,12};

       System.out.println("Before Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

       System.out.println();

         

       selectionsort(arr);

       

       System.out.println("After Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

   }

}  

Kindly check the second attached image below for the code output.

Bubble Sort in JAVA:-

public class bubblesort {

   static void bubblesort(int[] array) {

       int num = array.length;

       int temp = 0;

        for(int i=0; i < num; i++){

                for(int j=1; j < (num-i); j++){

                         if(array[j-1] > array[j]){

                           

                                temp = array[j-1];

                                array[j-1] = array[j];

                                array[j] = temp;

                        }

                         

                }

        }

   }

   public static void main(String[] args) {

               int arr1[] ={3333,60,25,32,55,620,85};

               

               System.out.println("Before Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

               System.out.println();

                 

               bubblesort(arr1);

               

               System.out.println("After Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

 

       }

}  

Kindly check the third attached image below for the code output.

Binary search in JAVA:-

public class binarysearch {

  public int binarySearch(int[] array, int x) {

     return binarySearch(array, x, 0, array.length - 1);

  }

  private int binarySearch(int[ ] arr, int x,

        int lw, int hg) {

     if (lw > hg) return -1;

     int middle = (lw + hg)/2;

     if (arr[middle] == x) return middle;

     else if (arr[middle] < x)

        return binarySearch(arr, x, middle+1, hg);

     else

        return binarySearch(arr, x, lw, middle-1);

  }

  public static void main(String[] args) {

     binarysearch obj = new binarysearch();

     int[] ar =

       { 22, 18,12,14,36,59,74,98,41,23,

        34,50,45,49,31,53,74,56,57,80,

        61,68,37,12,58,79,904,56,99};

     for (int i = 0; i < ar.length; i++)

        System.out.print(obj.binarySearch(ar,

           ar[i]) + " ");

     System.out.println();

     System.out.print(obj.binarySearch(ar,19) +" ");

     System.out.print(obj.binarySearch(ar,25)+" ");

     System.out.print(obj.binarySearch(ar,82)+" ");

     System.out.print(obj.binarySearch(ar,19)+" ");

     System.out.println();

  }

}

Kindly check the fourth attached image below for the code output

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