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lana [24]
3 years ago
11

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

l values in the vector (except the last value) with the last value in the vector subtracted from each value. Output each value on a new line. Ex: If the input is -46 66 76 9 -1, the output is:
-55
57
67
Computers and Technology
1 answer:
weqwewe [10]3 years ago
6 0

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

 nums.push_back(num);

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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Answer:

weightEarth = float(input("Enter weight on earth: "))

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Explanation:

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5 0
2 years ago
Which type of graph or chart will you use to show changes in data points?
likoan [24]
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6 0
3 years ago
Write a program that will predict the size of a population of organisms. The program should ask the user for the starting number
sineoko [7]

Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

 //Declaration of variables in the program

 double start_organisms;

 double daily_increase;

 int days;

 double updated_organisms;

 //The user enters the number of organisms as desired

 cout << "Enter the starting number of organisms: ";

 cin >> start_organisms;

 //Validating input data

 while (start_organisms < 2)

 {

     cout << "The starting number of organisms must be at least 2.\n";

     cout << "Enter the starting number of organisms: ";

     cin >> start_organisms;

 }

 //The user enters daily input, here's where we apply the 5.2% given in question

 cout << "Enter the daily population increase: ";

 cin>> daily_increase;

 //Validating the increase

 while (daily_increase < 0)

 {

     cout << "The average daily population increase must be a positive value.\n ";

     cout << "Enter the daily population increase: ";

     cin >> daily_increase;

 }

 //The user enters number of days

 cout << "Enter the number of days: ";

 cin >> days;

 //Validating the number of days

 while (days<1)

 {

     cout << "The number of days must be at least 1.\n";

     cout << "Enter the number of days: ";

     cin >> days;

 }

 

 //Final calculation and display of results based on formulas

 for (int i = 0; i < days; i++)

 {

     updated_organisms = start_organisms + (daily_increase*start_organisms);

     cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

     

     start_organisms = updated_organisms;

 }

 system("pause");

  return 0;

//end

}

Explanation:

6 0
3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
3 years ago
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