Answer:
#include<iostream>
int main()
{
int count=0;
int so_phan_tu;
std::cout << "nhap so phan tu : \n";
std::cin >> so_phan_tu;
int* A = new int[so_phan_tu];
std::cout << "nhap cac phan tu : \n";
for (int i = 0; i < so_phan_tu; i++)
{
std::cin >> A[i];
if (A[i] % 5 == 0)
{
count++;
}
}
std::cout << so_phan_tu - count;
system("pause");
return 0;
}
//cái này viết bằng C++ em nhé
Explanation:
Answer:
The Following are the solution to this question:
Explanation:
In Option a:
In the point (i) 
is transitive, which means it converts one action to others object because if 
indicates 
. It's true by definition, that becomes valid. But if 
, which implies 
. it's a very essential component. If 
. They 
will also be 
.
In point (ii), The value of 
is convergent since the 
. It means they should be dual a and b constant variable, therefore 
could only be valid for the constant variable, that is 
.
In Option b:
In this algorithm, the input size value is equal to 1 object, and the value of A is a polynomial-time complexity, which is similar to its outcome that is 
. It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop
. All internal loops operate on a total number of 
generations and therefore the final time complexity is .
Answer:
Encryption Keys
Explanation:
In order to make sure only the intended recipient receives the information, encryption keys rely on a unique pattern, just like your house key, except instead of grooves and ridges encryption keys use numbers and letters.
This would be copy and paste.
When you select a text, you can use your keys CTRL and C to copy that same selected text. After this, you can use the paste feature by selecting CTRL and V. Doing this, your selected text will be copied and pasted to a certain location.
