#include<stdio.h>
#include<conio.h>
void main()
{
int n, a[10, evensum=0, oddsum=0, i;
printf("\n Enter the number of values");
scanf("\n %d", &n);
printf("\n Enter %d number of values", n);
for(i=0; i<n; i++)
scanf("\n %d", &a[i]);
while(i<n)
{
if(a[i]%2==0)
evensum=evensum + a[i];
else
oddsum=oddsum + a[i];
}
printf("\n Odd sum = %d", oddsum);
printf("\n Even sum = %d", evensum);
getch();
}
Hopefully, is that correct, anyone out there.... I am a beginner and this is my homework, please help me correct the mistaks if you know them...
for(int i = 1; i < 100; i+=2) { System.out.println(i); }
The procedure is the same whether the number is even or odd. There is no separate procedure for odd numbers.
The answer is an odd number.
you cross the numbers out from back to front
15 and 13
for(int i = 1; i < 100; i+=2) { System.out.println(i); }
void f (int* a, int len) { if (!a) return;for (int i=0; i<len; ++i) { if (a[i]%2) { printf ("%d is odd\n", a[i]); } else { printf ("%d is even\n", a[i]); } }
int array[10] = {...}; for (int i = 0; i < 10; ++i) { if (i % 2 == 0) array[i] += 5; else array[i] -= 10; }
Reference: cprogramming-bd.com/c_page4.aspx#ODD%20 Numbers
The procedure is the same whether the number is even or odd. There is no separate procedure for odd numbers.
All nonzero numbers have factors. Some factors are odd numbers. 3 is an odd factor of 12.
Look for odd squares. Multiplying odd numbers results in an odd product.
#include
The answer is an odd number.
Shell problems are programs that can be run to find out information about numbers. The problem can help find an even or odd number, or what the sum of a cube is.
No. An odd plus an odd is even, and an even plus an odd is odd. Thus, you will always end up with an odd number if you add only three odd numbers together, and 50 is even.
you cross the numbers out from back to front