Please tell me the exact logic to solve this question. I’ve tried a lot but always end up with WA. Here is the link of the question - http://www.codechef.com/CDSU2015/problems/MHTBAG

It w1 is odd, the we need one n1 object to fill the last one space. If n1==0, then the point is wasted.

so, if w1 is odd, then subtract 1 from it (the last space), if n1>0, add 1 to answer and reduce one from n1. Do the same for w2.

Now you have two bags each with even number weight capacity. Now, each object in n2 carries 2 units of weight. So total weight is 2*n2.

Let, n2=2*n2;

and W=w1+w2;

So now, n2 and n1 are total weight available and W is the total weight capacity.

If W<=n2, it means we can fill all the weight in W by n2.

So, apply

{

ans+=W; //As W weight is available

n2-=W;

W=0;

}

else if(W>n2) // this means W can accommodate all n2 weight and will have spare space.

{

W-=n2;

ans+=n2;

n2=0;

}

Do the exact same for n1 and print the value in ans.

p.s. In solving with n2, if W becomes 0, it doesn’t cause any problem with n1 part as we are adding and subtracting 0 from any value.

**Easy to understand Solution**

```
#include<bits/stdc++.h>
using namespace std;
#define ll unsigned long long
int main(){
ll t, w1,w2,n1,n2,W;
cin>>t;
while(t--){
cin>>w1>>w2>>n1>>n2;
ll ans = 0;
ll temp = min(w1/2+w2/2,n2);
ans=2*temp;
ans += min(w1+w2-2*temp, n1);
cout<<ans;
}
return 0;
}
```