Just Even/Odd | Editorial by shivamjaiswal64

This is easy question . But its ***heavy I/O ***makes it partially accepted due to *TLE*.

So , we need to optimise our code.
Steps:
1) Take t input as no. of testcase.
2) In each testcase , Take n and q.
3) Declare three arrays as arr[] , even[] and odd[] with max value.
4) Count the number of even & odd from the arr[] and put the counted number in array of even[] and odd[] at ith index.
5) Take the Query. as l and r.
6) Subtract  the counted odd and even l from r. [  (even[r]-even[l-1])*(odd[r]-odd[l-1]) ]

For better understanding , Here is C code.

**Note**: This question might not be done by Java as  **Time Limit** is 1.0 sec(s) for each Heavy input file. 
Editorial by : [Shivam Jaiswal](https://www.stopstalk.com/user/profile/shivamjaiswal64)
```
#include<stdio.h>
#define max 200010
#define ll long long
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long int n,q,l,r;
        scanf("%lld %lld",&n,&q);
        long long int arr[max],even[max],odd[max];
        odd[0]=0;
        even[0]=0;
        for(long long int i=1;i<n+1;i++)
        {scanf("%lld",&arr[i]);
        if(arr[i]&1)
        {
            odd[i]=odd[i-1]+1;
            even[i]=even[i-1];
        }
        else{
            even[i]=even[i-1]+1;
            odd[i]=odd[i-1];
        }
        }
        while(q--){
            scanf("%lld%lld",&l,&r);
            l++;
            r++;
            ll res = (even[r]-even[l-1])*(odd[r]-odd[l-1]);
            printf("%lld\n",res);
        }
        
        
    }
}
```

Just Even/Odd | Editorial by shivamjaiswal64

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