Answer: 83.17
Explanation:
By definition, the dB is an adimensional unit, used to simplify calculations when numbers are either too big or too small, specially in telecommunications.
It applies specifically to power, and it is defined as follows:
P (dB) = 10 log₁₀ P₁ / P₂
Usually P₂ is a reference, for instance, if P₂ = 1 mW, dB is called as dBm (dB referred to 1 mW), but it is always adimensional.
In our question, we know that we have a numerical ratio, that is expressed in dB as 19.2 dB.
Applying the dB definition, we can write the following:
10 log₁₀ X = 19.2 ⇒ log₁₀ X = 19.2 / 10 = 1.92
Solving the logarithmic equation, we can compute X as follows:
X = 10^1.92 = 83.17
X = 83.17
Answer:
(a) What is the best case time complexity of the algorithm (assuming n > 1)?
Answer: O(1)
(b) What is the worst case time complexity of the algorithm?
Answer: O(n^4)
Explanation:
(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1)
.
(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).
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;
}
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