I've exceeded the time limit on Sherlock and Squares on hackerrank.com.

###Problem### Given the number of test cases and a range (inclusive), find the number of numbers within that range that are perfect squares.

###Sample Input###

• 2 (number of test cases)
• 3 9 (first test case)
• 17 24 (second test case)

###Sample Output###

• 2 (in range [3,9], there are two perfect square numbers)
• 0 (in range [17, 24], there are no perfect square numbers)

##My code in C++##

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange.

I've exceeded the time limit on Sherlock and Squares on hackerrank.com.

Problem

Given the number of test cases and a range (inclusive), find the number of numbers within that range that are perfect squares.

Sample Input

• 2 (number of test cases)
• 3 9 (first test case)
• 17 24 (second test case)

Sample Output

• 2 (in range [3,9], there are two perfect square numbers)
• 0 (in range [17, 24], there are no perfect square numbers)

My code in C++

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange.

###Problem### Given the number of test cases and a range (inclusive), find the number of numbers within that range that are perfect squares.

###Sample Input###

• 2 (number of test cases)
• 3 9 (first test case)
• 17 24 (second test case)

###Sample Output###

• 2 (in range [3,9], there are two perfect square numbers)
• 0 (in range [17, 24], there are no perfect square numbers)

##My code in C++##

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange. Thank you!

I've exceeded the time limit on Sherlock and Squares on hackerrank.com.

###Problem### Given the number of test cases and a range (inclusive), find the number of numbers within that range that are perfect squares.

###Sample Input###

• 2 (number of test cases)
• 3 9 (first test case)
• 17 24 (second test case)

###Sample Output###

• 2 (in range [3,9], there are two perfect square numbers)
• 0 (in range [17, 24], there are no perfect square numbers)

##My code in C++##

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange.

I am unsure of how I can simplify my code for this HackerRank "Sherlock and Squares" activity.

Here's my code in C++:

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange. Thank you!

###Problem### Given the number of test cases and a range (inclusive), find the number of numbers within that range that are perfect squares.

###Sample Input###

• 2 (number of test cases)
• 3 9 (first test case)
• 17 24 (second test case)

###Sample Output###

• 2 (in range [3,9], there are two perfect square numbers)
• 0 (in range [17, 24], there are no perfect square numbers)

##My code in C++##

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
 unsigned test_cases, low_range, high_range, count = 0;
 vector<unsigned> sherlock_squares;
 cin >> test_cases;
 while(test_cases != 0)
 {
 cin >> low_range >> high_range;
 for(unsigned i = low_range; i <= high_range; ++i)
 {
 if(sqrt(i) == floor(sqrt(i)))
 count += 1;
 }
 sherlock_squares.push_back(count);
 count = 0;
 --test_cases;
 }
 for(unsigned i = 0; i < sherlock_squares.size() - 1; ++i)
 cout << sherlock_squares.at(i) << endl;
 cout << sherlock_squares.at(sherlock_squares.size() - 1);
 return 0;
}

I'd appreciate it if an explanation is provided in your response, since I'm new to programming and Stack Exchange. Thank you!