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Elza [17]
3 years ago
14

Explain how it is possible for a sequence of packets transmitted through a wide area network to arrive at their destination in a

n order that differs from that in which they were sent. Why can’t this happen in a local network?)
Computers and Technology
1 answer:
dalvyx [7]3 years ago
5 0

Answer:

- Different route paths, due to dynamic routing in WAN.

- Local area networks have one or very few paths to destination and does not require dynamic routing.

Explanation:

A wide area network is a network that covers a large geographical area. It goes beyond the private local area network, with more routing paths and network intermediate devices. The router is an essential tool for routing packets between devices. It requires a routing path, learnt statically or dynamically to work.

There are mainly two types of route paths, they are, static routes and dynamic routes.

The dynamic routes are used mainly in WAN. Sometimes, there can be multiple path to a destination, the router determines the best path to send the packets. It sends the sequenced packets through all available path and they arrive at the destination depending on the path used, the packets arrive in an out -of-order fashion in the destination and a rearranged.

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Write a program num2rome.cpp that converts a positive integer into the Roman number system. The Roman number system has digits I
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Answer:

Explanation:

#include <stdio.h>  

int main(void)  

{    

   int num, rem;

   printf("Enter a number: ");

   scanf("%d", &num);

   printf("Roman numerals: ");        

   while(num != 0)

   {

       if (num >= 1000)       // 1000 - m

       {

          printf("m");

          num -= 1000;

       }

       else if (num >= 900)   // 900 -  cm

       {

          printf("cm");

          num -= 900;

       }        

       else if (num >= 500)   // 500 - d

       {            

          printf("d");

          num -= 500;

       }

       else if (num >= 400)   // 400 -  cd

       {

          printf("cd");

          num -= 400;

       }

       else if (num >= 100)   // 100 - c

       {

          printf("c");

          num -= 100;                        

       }

       else if (num >= 90)    // 90 - xc

       {

          printf("xc");

          num -= 90;                                              

       }

       else if (num >= 50)    // 50 - l

       {

          printf("l");

          num -= 50;                                                                      

       }

       else if (num >= 40)    // 40 - xl

       {

          printf("xl");            

          num -= 40;

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       else if (num >= 10)    // 10 - x

       {

          printf("x");

          num -= 10;            

       }

       else if (num >= 9)     // 9 - ix

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          printf("ix");

          num -= 9;                          

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          printf("iv");

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       }

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       {

          printf("i");

          num -= 1;                                                                                    

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   }

   return 0;

}

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Read integers from input and store each integer into a vector until -1 is read. Do not store -1 into the vector. Then, output al
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Answer:

The program in C++ is as follows:

#include <iostream>

#include <vector>

using namespace std;

int main(){

vector<int> nums;

int num;

cin>>num;

while(num != -1){

 nums.push_back(num);

 cin>>num; }  

for (auto i = nums.begin(); i != nums.end(); ++i){

    cout << *i <<endl; }

return 0;

}

Explanation:

This declares the vector

vector<int> nums;

This declares an integer variable for each input

int num;

This gets the first input

cin>>num;

This loop is repeated until user enters -1

while(num != -1){

Saves user input into the vector

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Get another input from the user

 cin>>num; }

The following iteration print the vector elements

<em> for (auto i = nums.begin(); i != nums.end(); ++i){ </em>

<em>     cout << *i <<endl; } </em>

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