There is the code below. You need to redo it using a function to check the parity of the entered number. After checking for parity, the program should ask for more numbers to enter.

# include & lt; iostream & gt;
using namespace std;
void main () {
  setlocale (LC_ALL, "ukr");
  int a;
  cout & lt; & lt; ("-Enter the number:");
  cin & gt; & gt; a;
  if (a% 2 == 0)
  {
    cout & lt; & lt; ("-This is an even number \ n");
  }
  else
  {
    cout & lt; & lt; ("-This is an Odd number \ n");
  }
  system ("pause");
}

Answer 1, authority 100%

Try it like this

std :: string str;
while (true)
{
  std :: cout & lt; & lt; ("-Enter the number:");
  std :: cin & gt; & gt; str;
  if (str == "quit")
    break;
  int value = :: strtol (str.c_str (), 0, 10);
  if (value% 2 == 0)
    std :: cout & lt; & lt; ("-number \ n");
  else
    std :: cout & lt; & lt; ("-Odd \ n");
}

Only here the situation is not handled when an invalid string is transmitted

Answer 2, authority 50%

for (;;)
{
  int N;
  cout & lt; & lt; "Enter a number; 0 to complete:";
  if (! (cin & gt; & gt; N) || (N == 0)) break;
  cout & lt; & lt; N & lt; & lt; "-" & lt; & lt; ((N & amp; 1)? "Not": "") & lt; & lt; "even number \ n";
}

Answer 3

You would add a loop

for (i = 1; i & lt; = 1000; i ++)
  {
  cout & lt; & lt; ("-Enter the number:");
  cin & gt; & gt; a;
  if (a% 2 == 0)
{
  cout & lt; & lt; ("-This is an even number \ n");
}
else
{
cout & lt; & lt; ("-This is an Odd number \ n");
}
  }

then the loop will check 1000 times, it all depends on how much and under what conditions to check for parity, otherwise another loop may be needed