Read in even numbers (c), and some of the odd number of integers for the end of the language

Read in even numbers (c), and some of the odd number of integers for the end of the language

#include
main()
{
int a,b,c;
b=0;c=0;
while(1) {
scanf("%d",&a);
if (!a) break;
if (a%2) b++;
else c++;
}
Printf ("odd =% D, even =% D, n", B, c));
}

Solve C language input a positive integer, judge whether the number is odd or even
Enter a positive integer to determine whether the number is odd or even

#Include & nbsp; & lt; stdio. H & gt; void & nbsp; main() {& nbsp; & nbsp; & nbsp; & nbsp; int & nbsp; n; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; printf & quot; please enter a positive integer: n & quot;; & nbsp; & nbsp; & nbsp; scanf & quot;% D & quot;, & nbsp; & amp; n

Using C language, how to program and calculate any n integers, count the odd sum, odd number, even number and even number respectively

#Include void main() {int n, I, Ji = 0, Ou = 0; / / Ji is used to count the number of odd numbers, Ou is used to count the number of even numbers, int input, Jihe = 0, Ouhe = 0; / / Jihe is used to count the sum of odd numbers, Ouhe is used to count the sum of even numbers, input is the input number scanf ("% D / N", & n); / / input number n for (I = 0; I

In this paper, we define an operation Θ, when m and N are both positive even numbers or positive odd numbers, m Θ n = m + N, when one of M and N is positive odd number and the other is positive even number,
Then the number of elements in the set M = {(a, b) a Θ B = 16, a ∈ n *, B ∈ n *} is?

One odd one even (1,16) (16,1) two kinds
Tongqi (1,15) (3,13) (5,11) (7,9) 4x2 = 8 species
Homozygous (2,14) (4,12) (6,10) (8,8) 3x2 + 1 = 7 species
There are 17 kinds in total

This paper defines a kind of operation ⊙ when m and N are both positive even numbers or positive odd numbers, m ⊙ n = m + n
When one of M and N is positive odd and the other is positive even, m ⊙ n = Mn
Then the number of elements in the set M = {(a, b) | a ⊙ B = 12, a, B ∈ n *} is ()

It can be written directly
1 ° m and N are positive even numbers or positive odd numbers
(1) M = 1, n = 11, (2) M = 2, n = 10
When one of 2 ° m and N is positive odd number and the other is positive even number
(1) M = 1, n = 12, (2) M = 3, n = 4,4,3,12,1, a total of four
To sum up, there are 16

If n is an integer greater than 1, is the value of power n of P = n + (n times n-1) multiplied by 1 / 2 - (- 1) necessarily even or odd?

P=n+[n*(n-1)]/2-(-1)↑n=[n-(-1)↑n]+[n*(n-1)]/2
1、 If n is odd, then [n - (- 1) ↑ n] must be even; [n * (n-1)] / 2 is odd even indeterminate
2、 If n is even, then [n - (- 1) ↑ n] must be odd; [n * (n-1)] / 2 is odd even indeterminate
For reference only

If n is an integer greater than 1, is the value of the power n of P = n + (n-1) * 1 / 2 - (- 1) even or odd, or both?
I hope I can be more careful

If n is odd, it is impossible to judge whether the value of P is odd or even, because when (n-1) * 1 / 2 is odd, P is odd, and when (n-1) * 1 / 2 is even, P is even. Then both are possible. Therefore, when n is odd, the answer is both possible

If n is an integer greater than 1, is the value of the power n of P = n + (n-1) multiplied by 1 / 2 - (- 1) necessarily even or odd?
reason

1 - (- 1) ^ n = 0 when n is even
The original formula = n is even
1 - (- 1) ^ 2 = 2 when n is odd
The original formula = n + n-1 = 2N-1 is odd

Mathematics problem of grade one in junior high school: n is an integer greater than 1, find the value of power n of (- 6) + power N-1 of power n of (- 6) (hint: n is odd or even)

N is odd
Power n of (- 6) + power N-1 of (- 6)
=-6^n+6*6^(n-1)
=-6^n+6^n=0
N is even
Power n of (- 6) + power N-1 of (- 6)
=6^n-6*6^(n-1)
=6^n-6^n=0

Given a < 0, if the value of - 3A ^ n · A & # is greater than zero, then the value of N can only be () a.n is odd, B.N is even, C.N is positive integer, D.N is integer

If the value of - 3A ^ n &; a ^ 2 is greater than zero, - 3A ^ n &; a ^ 2 = - 3A ^ (n + 2) > 0, a ^ (n + 2) < 0,
But a < 0
N + 2 is odd,
So n is odd
So choose a