THE bestValue PROBLEM Using the Camera structure defined in file p1.cpp, write the function named bestValue(). The function takes one input parameter: a vector of Camera. The vector is sorted by manufacturer, so all cameras of a particular manufacturer are together in the vector. The function returns a vector of string, with the manufacturer, model number and price concatenated together exactly in this format.

#include <iostream>

#include <vector>

#include <string>

#include <cmath>

#include <iomanip>

#include <sstream>

using namespace std;

// Given struct

struct Camera {

// Declaring variables

string manufacturer;

string model;

int releaseYear;

int resolution;

int weight;

double price;

};

vector<string> bestValue(const vector<Camera>& cameraNew) {

vector<string> display;

double winnerOne = 0;

double winnerTwo = 0;

double realWinnerOne = 0;

double realWinnerTwo = 0;

string displayOne;

string displayTwo;

string ans;

int find = 0;

// Calculating best value using given formula

for (unsigned i = 1; i < cameraNew.size(); i++) {

winnerOne = cameraNew[i].price / static_cast<double>(cameraNew[i].resolution);

winnerTwo = cameraNew[i-1].price / static_cast<double>(cameraNew[i - 1].resolution);

if (cameraNew[i].manufacturer == cameraNew[i - 1].manufacturer) {

if (winnerOne > winnerTwo && winnerOne > realWinnerOne) {

find = 1;

}

else if (winnerOne < winnerTwo && winnerTwo > realWinnerTwo) {

find = 2;

}

if (find == 1) {

realWinnerOne = cameraNew[i].price/ static_cast<double>(cameraNew[i].resolution) ;

ostringstream displayStream1;

displayStream1 << fixed << setprecision(2) << cameraNew[i].price;

displayOne = cameraNew[i].manufacturer + ":" + cameraNew[i].model + ":$" + displayStream1.str();

find = 0;

}

else if (find == 2) {

realWinnerTwo = cameraNew[i - 1].price / static_cast<double>(cameraNew[i - 1].resolution);

ostringstream displayStream2;

displayStream2 << fixed << setprecision(2) << cameraNew[i - 1].price;

displayTwo = cameraNew[i-1].manufacturer + ":" + cameraNew[i-1].model + ":$" + displayStream2.str();

find = 0;

}

}

else {

if (realWinnerOne >= realWinnerTwo) {

display.push_back(displayOne);

realWinnerOne = 0;

realWinnerTwo = 0;

}

else if (realWinnerOne < realWinnerTwo) {

display.push_back(displayTwo);

realWinnerOne = 0;

realWinnerTwo = 0;

}

}

}

if (realWinnerOne > realWinnerTwo) {

display.push_back(displayOne);

}

else if (realWinnerOne < realWinnerTwo) {

display.push_back(displayTwo);

}

else if (realWinnerOne == realWinnerTwo) {

display.push_back(displayTwo);

}

return display;

}

template <typename T>

ostream& operator<<(ostream& out, const vector<T>& display) {

if (display.size() > 0) {

out << "0. " << display[0];

out << endl;

//displaying output in the given format

for (size_t i = 1; i < display.size(); i++) {

out << i << ". " << display[i];

out << endl;

}

}

else

{

out << "Size of the vector is 0 or less than 0." << endl;

}

return out;

}

void camBestValue() {

const vector<Camera> vCameras = {

{ "Agfa", "ePhoto 1280", 1996, 1024, 400, 180 },

{ "Agfa", "ePhoto CL45", 2000, 1600, 275, 180 },

{ "Canon", "PowerShot 350", 1996, 640, 315, 150 },

{ "Canon", "PowerShot 600", 1994, 832, 445, 139 },

{ "Canon", "PowerShot A10", 2001, 1280, 355, 139 },

{ "Casio", "Exilim EX-P505", 2005, 2560, 250, 260 },

{ "Casio", "Exilim EX-P600", 2006, 2816, 275, 260 },

{ "Casio", "Exilim EX-P700", 2006, 3072, 275, 260 },

{ "Epson", "PhotoPC 800", 1997, 1600, 285, 220 },

{ "Epson", "PhotoPC L-500V", 2004, 2560, 205, 150 },

{ "Fujifilm", "FinePix 40i", 2000, 2400, 185, 180 },

{ "Fujifilm", "FinePix 50i", 2001, 2400, 205, 180 },

{ "Fujifilm", "DS-260HD", 1997, 1280, 845, 190 },

{ "Fujifilm", "DS-300", 1995, 1280, 845, 200 },

{ "HP", "Photosmart 320", 2002, 1632, 230, 190 },

{ "HP", "Photosmart 435", 2003, 2048, 180, 190 },

{ "HP", "Photosmart 620", 2002, 1632, 260, 190 },

};

vector<string> display;

display = bestValue(vCameras);

cout << display << endl;

}

int main() {

camBestValue();

}

Explanation:

The program takes one input parameter: a vector of Camera. The vector is sorted by manufacturer, so all cameras of a particular manufacturer are together in the vector.

The function returns a vector of string, with the manufacturer, model number and price concatenated together.

It produces the best value.

i believe the answer would be b. within an intersection or on a crosswalk. hope this ! : )

answer; i believe the correct answer is network?