MechanicalCounter | Topcoder Archive

Problem Statement

You work for a company that sells mechanical bit counting devices. These devices wear out when they do a lot of counting. You are going to see how much wear has been put on one particular 15-bit binary counter. One unit of wear occurs for every bit that is flipped in the counter. For example:

If the counter was at     7 = 000000000000111 
And it was incremented to 8 = 000000000001000

Then 4 bits are flipped so 4 units of wear would occur. If a counter was incremented through an inclusive range like 7 -> 12 then :

Starts at 7 = 000000000000111 
Goes to   8 = 000000000001000   7 -> 8 = 4 units of wear
Goes to   9 = 000000000001001   8 -> 9 = 1 unit of wear
Goes to  10 = 000000000001010   9 -> 10 = 2 units of wear
Goes to  11 = 000000000001011   10 -> 11 = 1 unit of wear
Goes to  12 = 000000000001100   11 -> 12 = 3 units of wear

This causes a total of 4+1+2+1+3 = 11 units of wear. Return the total wear caused by incrementing the counter through the inclusive range between start and finish.

Definition

Class:: MechanicalCounter
Method:: amountWear
Parameters:: int, int
Returns:: int
Method signature:: int amountWear(int start, int finish)
(be sure your method is public)

Constraints

start will be less than finish.
start will be between 0 and 32766 inclusive.
finish will be between 1 and 32767 inclusive.

Examples

```
7
```
```
8
```
```
Returns: 4
```
As seen above, 4 units of wear occur when 7 becomes 8.
```
7
```
```
12
```
```
Returns: 11
```
The example in the problem statement.
```
1
```
```
32767
```
```
Returns: 65518
```

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