After scanning all array elements how can i printf even and odd seperately.

```
#include <stdio.h>
void main()
{
int arr[100];
int n,i;
printf("Number of Elements:-");
scanf("%d",&n);
for (i=0;i<=n;i=i+1)
{
printf("Elem %d:-",i);
scanf("%d",&arr[i]);
}
for(i=0;i<=n;i=i+1)
{
if(arr[i]/2==0)
{
printf("Even%d\n",arr[i]);
}
else
{
printf("odd%d\n",arr[i]);
}
}
}
```

Output:

Number of Elements:-4 Elem 0:-1 Elem 1:-2 Elem 2:-3 Elem 3:-4 Elem 4:-5 Even1 odd2 odd3 odd4 odd5

### >Solution :

The conventional way to write your loop is:

```
for( i=0; i<n; ++i )
{
printf("%s %d\n", (arr[i]%2)? "Odd" : "Even", arr[i]);
}
```

In C, most loops have an initial condition of `var=0`

, run while `var<target`

, and increment with `++var`

(some people prefer `var++`

)

The conventional way to test if a number is even or odd is using the Modulo (`%`

) operator, not using division (`/`

).

Why do you think a an expression like `10/2`

would ever equal `0`

??

`10`

is clearly even, dividing by `2`

gives `5`

, not zero!!?

Putting it all together, I got:

```
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int arr[100];
int n = 20;
for ( int i=0; i<n; ++i )
{
arr[i] = rand() % 100;
// Replaced user-input with pseudo-random numbers, for ease and simplicity.
// You can stick with scanf if you want.
}
for( int i=0; i<n; ++i )
{
printf("%-4s %d\n", (arr[i]%2)? "Odd" : "Even", arr[i]);
}
return 0;
}
```

## Output:

```
Success #stdin #stdout 0.01s 5404KB
Odd 83
Even 86
Odd 77
Odd 15
Odd 93
Odd 35
Even 86
Even 92
Odd 49
Odd 21
Even 62
Odd 27
Even 90
Odd 59
Odd 63
Even 26
Even 40
Even 26
Even 72
Even 36
```