________ are the most popular method used by visitors to find web sites.
1 answer:
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#include<iostream>
#include <fstream>
#include <vector>
#include <string>
#include "StatePair.h"
using namespace std;
int main() {
ifstream inFS;
int zip;
int population;
string abbrev;
string state;
unsigned int i;
vector<StatePair <int, string>> zipCodeState;
vector<StatePair<string, string>> abbrevState;
vector<StatePair<string, int>> statePopulation;
inFS.open("zip_code_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file zip_code_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <int, string> temp;
inFS >> zip;
if (!inFS.fail()) {
temp.SetKey(zip);
}
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetValue(abbrev);
}
zipCodeState.push_back(temp);
}
inFS.close();
inFS.open("abbreviation_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file abbreviation_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, string> temp;
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetKey(abbrev);
}
getline(inFS, state);
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetValue(state);
}
abbrevState.push_back(temp);
}
inFS.close();
inFS.open("state_population.txt");
if (!inFS.is_open()) {
cout << "Could not open file state_population.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, int> temp;
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetKey(state);
}
inFS >> population;
if (!inFS.fail()) {
temp.SetValue(population);
}
getline(inFS, state);
statePopulation.push_back(temp);
}
inFS.close();
cin >> zip;
for (i = 0; i < zipCodeState.size(); ++i) {
}
for (i = 0; i < abbrevState.size(); ++i) {
}
for (i = 0; i < statePopulation.size(); ++i) {
}
}
Statepair.h
#ifndef STATEPAIR
#define STATEPAIR
#include <iostream>
using namespace std;
template<typename T1, typename T2>
class StatePair {
private:
T1 key;
T2 value;
public:
StatePair()
{}
void SetKey(T1 key)
{
this->key = key;
}
void SetValue(T2 value)
{
this->value = value;
}
T1 GetKey() { return key;}
T2 GetValue() { return value;}
void PrintInfo()
{
cout<<key<<" : "<<value<<endl;
}
};
Answer:
8% of the picture
Explanation:
Given:
Square picture pixels = 25 MP
Pixels that can be displayed by the computer = 2000 pixels by 1000 pixels
or
Pixels that can be displayed by the computer = 2000000 pixels
Now,
The fraction of picture viewable on the screen =
or
The fraction of picture viewable on the screen = 0.08
or
The fraction of picture viewable on the screen = 8% of the picture