0
#include
using std::cout;
using std::endl;
int main()
{
int arr[] = {2, 45, 34, 56, 56};//you define here your array of even and odd numbers
int arrSize = sizeof arr/sizeof arr[0];
int numEven = 0;
int numOdd = 0;
for (int i = 0; i < (arrSize - 1); i++)
{
if (arr[i] % 2 == 0)
{
numEven++;
}
else
{
numOdd++;
}
}
cout << endl << "Number of even numbers: " << numEven
<< endl << "Number of odd numbers: " << numOdd
<< endl;
system("PAUSE");
return 0;
}
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How do you find odd or even in n number of value?
-- First of all, if the number is a fraction or a decimal, then don't worry about it. Only whole numbers are considered odd or even. -- All multiples of 2 are even numbers, and that's easy to recognize. It doesn't matter how big the number is, or how many digits it has. You only have to look at the last digit (the one on the right-hand end). If that digit is 2, 4, 6, 8, or 0, then the whole big number is even. Otherwise it's odd.
How do you Write a C program to print a Series of Odd Numbers and Even Numbers.?
Reference:http:cprogramming-bd.com/c_page2.aspx# strange number
What is odd loop?
Move the print out requesting the user to enter an integer outside of the for loop and it will only print once instead of each time around the loop. You'll need a way to save the even and odd numbers that you detect in the loop. One way is to have separate arrays to hold the even and the odd numbers as you go around the loop. Then at the end of the loop you can have more loops to print the contents of one array and then the contents of the other array. Another way is to concatenate the number onto separate Strings (even and odd) to be displayed after the data gathering loop.
Without using mod operator find the even or odd?
import java.util.*; class Demo { public static void main(String args[]) { System.out.println("Even Numbers Are "); for (int i=1;i<=20;i++) { System.out.print(" "+(2*i)); } System.out.println("\n \n Odd Numbers Are "); for(int j=0;j<=20;j++) { System.out.print(" "+((2*j)+1)); } } }
Write an algorithm find out the sum of first n odd numbers?
Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the open range (1:n)sum := 0val := 2while (val < n) do{sum := sum + valval := val + 2}return sumNote that you explicitly specified between 1 and n, which literally means both 1 and n should be excluded from the sum. 1 would be excluded anyway since it is not an even number, however if we wish to include n, then use the following algorithm instead:Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the half-open range (1:n]sum := 0val := 2while (val
