For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even
odd loop means at least the loop execute once.
By learning how to program on C+.
int i, sum = 0; for (i=0; i<20; i+=2) sum+=i;
Even (unless c = 0 in which case it is either or both!)
To write a C program to determine if something is odd or even you need to be a programmer. To write a program in C is complicate and only done by programmers.
Yes. If n is odd, then n + c where c is an even constant will be odd. n + d where d is an odd constant will be even.
for (int i=1; i<20; i+=2) printf ("%d\n", i);
for (int i = 15; i < 30; i += 2) cout << i << endl;
no, b/c you can think of an odd number as an even number plus one. So if you add two odd numbers, you would be adding 2 even numbers plus 2... Let # be even... (#+1) + (#+1) = (#+#) + 2 two even numbers added together are always even
Yes, you can program games with C++.
write a program in C that prompts the user with the following lines: a) Add two integers c) Compare two integers for the larger t) Test an integers for odd or even q) Quit
#include <stdio.h> main() { int n, odd=0, even=0; while (scanf("%d",&n)&&(n!=0)) (n%2)?++odd:++even; printf("odd: %d even: %d\n",odd,even); }
For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even
Exit the program and relaunch it.
odd loop means at least the loop execute once.
#include <iostream> using namespace std; int main() { int x; cout << "Enter a number: "; cin >> x; if(x % 2 == 0) { cout << x << " is even" << endl; } else { cout << x << " is odd" << endl; } char wait; cin >> wait; return 0; }