CareerCup

#include<stdio.h>
#include<conio.h>
void main()
{
int a[10],cnt=0,q;
printf("Enter the Length");
scanf_s("%d", &q);
printf("Enter the nos in the array \n");
for (int i = 0; i < q; i++)
{
printf("\n");
scanf_s("%d", &a[i]);

}

for (int j = 0; j < q; j++)
{
for (int k = j + 1; k < q; k++)
{
if ((a[j] + a[k]) % 2 == 0)

{
cnt += 1;

}
}

}
printf("\n%d\n", cnt);
_getch();
}

Give an example,please.

Its actually pretty straight forward. Even sum can either be obtained by adding 2 odd or 2 even numbers. So determine how many even and odd number in the array. That is simple.

And then to find total even sums of two numbers, all you need is a simple formula.

package com.google.practice;

public class EvenSum {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		int[] arr = {1,2,5,3,6,7,1,4,0,-4,1,-2,1,0,1};

		int oddCount=0,evenCount=0;
		
		for(int i=0;i<arr.length;i++){
			if(arr[i]%2==0)
					evenCount++;
			else
					oddCount++;
		}	
		
		System.out.println(((evenCount-1)*(evenCount)/2)+((oddCount-1)*(oddCount)/2));
	}

}