You just need a little bit of math to solve this. Add up the items and see which one would be equal to $55.00 or less.
A = $59.90 so this is not the answer
B = $60.94 so this is not the answer
C = $55.99 so this is not the answer
D = $50.97
D is the correct answer.
Before cellphones are able to be used to browse the internet, play games, record videos and take photos, its main purpose is similar to a telephone, albeit it is more portable. Early cellphones would not have games for you to play, not would it have emails for you to check, read, and reply to. Though ringtones might seem as an acceptable option, early cellphones would also have many selections or even any for you to choose.
Thus, the best option would be (B) contact list, which is necessary for a cellphone to have since the owner would need the number to be able to make a call.
Hi,
I changed your program using some of the concepts you were trying to use. Hopefully you can see how it works:
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
short T;
cin >> T;
cin.ignore();
string str[100];
for(int i=0; i<T; i++)
{
getline(cin, str[i]);
}
for (int i = 0; i < T; i++)
{
stringstream ss(str[i]);
string tmp;
vector<string> v;
while (ss >> tmp)
{
// Let's capitalize it before storing in the vector
if (!tmp.empty())
{
transform(begin(tmp), end(tmp), std::begin(tmp), ::tolower);
tmp[0] = toupper(tmp[0]);
}
v.push_back(tmp);
}
if (v.size() == 1)
{
cout << v[0] << endl;
}
else if (v.size() == 2)
{
cout << v[0][0] << ". " << v[1] << endl;
}
else
{
cout << v[0][0] << ". " << v[1][0] << ". " << v[2] << endl;
}
}
return 0;
}
