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Sidana [21]
3 years ago
9

Write a program that first gets a list of integers from input. The input begins with an integer indicating the number of integer

s that follow. That list is followed by two more integers representing lower and upper bounds of a range. Your program should output all integers from the list that are within that range (inclusive of the bounds). For coding simplicity, follow each output integer by a space, even the last one. The output ends with a newline.
Ex: If the input is: 5 25 51 0 200 33
0 50

then the output is:
25 0 33
Computers and Technology
1 answer:
snow_tiger [21]3 years ago
3 0

Answer:

The program implemented in C++ is as follows:

#include <iostream>

#include <vector>

using namespace std;

int main(){

int lenlist;

vector<int> mylist;

cout<<"Length List: ";

cin>>lenlist;

mylist.push_back(lenlist);

int i =0; int num;

while(i<lenlist){

   cin>>num;

   mylist.push_back(num);  

   i++;

}

int min,max;

cout<<"Min: ";  cin>>min;

cout<<"Max: ";  cin>>max;

cout<<"Output!"<<endl;

for(int i=1; i < mylist.size(); i++){

  if(mylist.at(i)>=min && mylist.at(i)<=max){

      cout<<mylist.at(i)<<" ";

  }

}

  return 0;

}

Explanation:

This declares the length of the list as integer

int lenlist;

This declares an integer vector

vector<int> mylist;

This prompts user for length of the list

cout<<"Length List: ";

This gets input of length of the list from the user

cin>>lenlist;

This pushes user input to the vector

mylist.push_back(lenlist);

int i =0; int num;

The following iteration gets user inputs and pushes them into the vector

<em>while(i<lenlist){ </em>

<em>    cin>>num; </em>

<em>    mylist.push_back(num);  </em>

<em>    i++; </em>

<em>} </em>

This declares min and max variables

int min,max;

This prompts user for the lower bound

cout<<"Min: ";  cin>>min;

This prompts user for the upper bound

cout<<"Max: ";  cin>>max;

The following iteration checks for numbers within the lower and upper bound and print the numbers in that range

<em>cout<<"Output!"<<endl; </em>

<em>for(int i=1; i < mylist.size(); i++){ </em>

<em>   if(mylist.at(i)>=min && mylist.at(i)<=max){ </em>

<em>       cout<<mylist.at(i)<<" "; </em>

<em>   } </em>

<em>} </em>

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Answer:

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Explanation:

E = integer value of exponent

M = significand value

F = fractional value

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A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

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