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Bad White [126]
2 years ago
9

Implement the sum_positive_numbers function, as a recursive function that returns the sum of all positive numbers between the nu

mber n received and 1. For example, when n is 3 it should return 1+2+3=6, and when n is 5 it should return 1+2+3+4+5=15.
Computers and Technology
1 answer:
lakkis [162]2 years ago
5 0

Answer:

def sum_positive_numbers(n):

if(n==0 or n==1): #if n is 0 or 1 return n as its sum also will be n

return n

return n+sum_positive_numbers(n-1) #else return n and sum of all the number smaller then n

print(sum_positive_numbers(3)) # Should be 6

print(sum_positive_numbers(5)) # Should be 15

Explanation:

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Answer:

The code is given below with appropriate comments

Explanation:

// TestSolution class implementation

import java.util.Arrays;

public class TestSolution

{

  // solution function implementation

  public static int solution(int[] arr)

  {

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      int i, j, count = 0;

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      // use the nested loops to count the number of moments for which every turned on bulb shines

      for (i = 0; i < arr.length; i++)

      {

          shines = true;

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          {

              if (arr[i] > arr[j])

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          }

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              count++;

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  } // end of solution function

 

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      int[] D = new int[N];

     

      // fill the array D with the distinct random numbers within the range [1..N]

      int i = 0;

      while(i < N)

      {

          int num = 1 + (int)(Math.random() * N);          

          boolean found = false;

          for(int j = 0; j < i && !found; j++)

          {

              if(D[j] == num)

                  found = true;

          }

         

          if(!found)

          {

              D[i] = num;

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      // print the elements and number of moments of the arrays A, B, and C

      System.out.println("Array A: " + Arrays.toString(A) + " and Moments: " + solution(A));

      System.out.println("Array B: " + Arrays.toString(B) + " and Moments: " + solution(B));

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  } // end of main function

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Answer:

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2 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

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Explanation:

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2 years ago
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