Answer:
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
double num;
// read discriminant
cin>>num;
// calculate square of the input
double result=num*num;
// print the square number
cout<<result<<endl;
return 0;
}
Explanation:
Declare a variable "num".Read the value of "num" from keybord.Then calculate square of the input number.Print the square of input number.
Output:
5
25
Answer:
B. {1, 2, 2, 3, 3, 4, 5}
Explanation:
Given
The above code segment
Required
Determine which list does not work
The list that didn't work is 
Considering options (A) to (E), we notice that only list B has consecutive duplicate numbers i.e. 2,2 and 3,3
All other list do not have consecutive duplicate numbers
Option B can be represented as:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 2](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%202)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 3](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%203)
![nums[5] = 4](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%204)
![nums[6] = 5](https://tex.z-dn.net/?f=nums%5B6%5D%20%3D%205)
if (nums.get(j).equals(nums.get(j + 1)))
The above if condition checks for duplicate numbers.
In (B), when the elements at index 1 and 2 (i.e. 2 and 2) are compared, one of the 2's is removed and the Arraylist becomes:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 3](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%203)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 4](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%204)
![nums[5] = 5](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%205)
The next comparison is: index 3 and 4. Meaning that comparison of index 2 and 3 has been skipped.
<em>This is so because of the way the if statement is constructed.</em>