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3 years ago

Write a short java method that takes an integer n and returns the sum of all the odd positive integers less than or equal to n.

1 answer:

3 0

Public int add2n( int n )
{
    int total;
    if( n % 2 == 0 )
        total = n--;

    while( n > 0 )
    {
        total += n;
        n -= 2;
   }
   return( total );
}

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Which principle suggests that specific single responsibility interfaces are better than one general purpose interface?

ElenaW [278]

Answer:Interface segregation principle

Explanation: Interface-segregation principle (ISP) is amongst the major principles of the object-oriented design which describes that none of the users/clients can be forced for indulging and depending on the unknown methods or methods that they don't have knowledge about.

It functions by making the interfaces visible to the user that specifically fascinates them and keeping other smaller interfaces.Interfaces are made by splitting process and making the small interfaces from them.

8 0

3 years ago

Write a loop statement to count and display the number of positive integers in the array. That is, your loop should display the

dlinn [17]

Answer:

int count =0;

for(int i=0;i<10;i++)

{

if(myArray[i]>=0)

{

count++;

}

cout<<"Number of positive integers is "<<count<<endl;

Explanation:

The above written loop is for counting positive integers in the myArray[].

For counting we have taken a count integer initialized with 0.On iterating over the array if the element is greater than or equal to 0 we consider it as positive and increasing the count.At the end printing the count.

3 0

3 years ago

Consider a set of mobile computing clients in a certain town who each

poizon [28]

Answer: answer given in the explanation

Explanation:

We have n clients and k-base stations, say each client has to be connected to a base station that is located at a distance say 'r'. now the base stations doesn't have allocation for more than L clients.

To begin, let us produce a network which consists of edges and vertex

Network (N) = (V,E)

where V = [S, cl-l, - - - - cl-n, bs-l - - - - - - bs-k, t]

given that cl-l, - - - - - cl-n represents nodes for the clients

also we have that bs-l, - - - - - bs-k represents the nodes for base station

Also

E = [ (s, cl-i), (cl-i,bs-j), (bs-j,t)]

(s, cl-i) = have capacity for all cl-i (clients)

(cl-i,bs-j) = have capacity for all cl-i clients & bs-j (stations)

⇒ using Fond Fulkorson algorithm we find the max flow in N

⇒ connecting cl-i clients to bs-j stations

like (cl-i, bs-j) = 1

if f(cl-i, bs-j) = 0

⇒ say any connection were to produce a valid flow, then

if cl-i (clients) connected f(s,cl-i) = 1 (o otherwise)

if cl-i (clients) connected to bs-j(stations) f(cl-i,bs-j) = 1 (o otherwise)

f(bs-j,t) = no of clients (cl-i) connected to bs-j

⇒ on each node, the max flow value (f) is longer than the no of clients that can be connected.

⇒ create the connection between the client and base station i.e. cl-l to base bs-j iff f(cl-i, bs-j) = 1

⇒ when considering the capacity, we see that any client cannot directly connect to the base stations, and also the base stations cannot handle more than L clients, that is because the load allocated to the base statsion is L.

from this, we say f is the max no of clients (cl-i) that can be connected if we find the max flow, we can thus connect the client to the base stations easily.

cheers i hope this helps

5 0

3 years ago

HELP I WILL MARK BRAINLIEST!!! I NEED ASAP!!!

Alex777 [14]

#This is a way without a loop

friends = list(map(str,input("Enter Names: ").split()))

print(sorted(friends))

#This is a way with a loop (for&&while)

friends = list(map(str,input("Enter Names: ").split()))

cool = True

while cool:

cool = False

for i in range(len(friends)-1):

if friends[i] > friends[i+1]:

coo = friends[i]

friends[i] = friends[i+1]

friends[i+1] = coo

cool = True

print(friends)

4 0

3 years ago

I am trying to code a lifting simulator game on ro-blox. These problems listed in the third picture tells you the problems. If y

ipn [44]

Answer:

cool

Explanation:

3 0

2 years ago

Read 2 more answers