Answer / nikhil patil

#include<stdio.h>
#include<conio.h>main()
{
int a[5],even=0,odd=1,i;

for(i=0;i<5;i++)

scanf("%d",&a[i]);
/* display the number of odd and even intergers */
for(i=0;i<5;i++)
{
if((a[i]%2 ==0))
even++;
if((a[i]%2==1))
odd++;
}
printf("%d",even,);
printf ("%d",odd,);
getch();

Answer / april

#include<stdio.h>
#include<conio.h>

using namespace std;
int main();
{
int i,j;
int num[20],i,odd=0,even=0,n;
cout<<("enter the size of the array",n);
cin>>("%d",&n);

cout<<("Enter the %d array elements",n);
for(i=20;i<n;i++);
{
cout<<("%d",&num[i]);
}
for(i=0;i<n;i++);
{
if(num[i]%2==0);
{
even++;
}
else
{
odd++;
}
}
cout<<("\n The number of even numbers in the array :%
d ",even);
cout<<("\n The number of odd numbers in the array : %
d",odd);
getch();
system("pause");
}

Answer / jasna.c

#include<stdio.h>
#include<conio.h>
void main()
{
int arr[100],i,odd=0,even=0,n;
printf("enter the size of the array");
scanf("%d",&n);
printf("Enter the %d array elements",n);
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
for(i=0;i<n;i++)
{
if(arr[i]%2==0)
{
even++;
}
else
{
odd++;
}
}
printf("\n The number of even numbers in the array :%d ",even);
printf("\n The number of odd numbers in the array : %d",odd);
getch();
}

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int *a[5] refers to

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