NumReverseEasy | Topcoder Archive

Problem Statement

Take all the positive integers from A to B, inclusive.

Reverse as many of them as you like. (E.g., reversing 1234 changes it into 4321, and reversing 4700 turns it into 0074 = 74.)

Then, add them all up.

Calculate and return the largest possible result you can get.

Definition

Class:: NumReverseEasy
Method:: getsum
Parameters:: int, int
Returns:: long
Method signature:: long getsum(int A, int B)
(be sure your method is public)

Constraints

B will be between 1 and 100,000, inclusive.
A will be between 1 and B, inclusive.

Examples

```
21
```
```
23
```
```
Returns: 75
```
We have the numbers 21, 22, and 23. We can reverse 23 to get 32. This will give us the final sum 21 + 22 + 32 = 75, which is the largest result we can get from these three numbers.
```
12
```
```
21
```
```
Returns: 489
```
Note that after we reverse 12, our collection of numbers will contain two separate 21s. Each of them contributes to the final sum.
```
97
```
```
101
```
```
Returns: 495
```
Here an optimal solution is not to reverse anything.
```
123
```
```
127
```
```
Returns: 2605
```
Here an optimal strategy is to reverse everything.
```
1
```
```
100000
```
```
Returns: 6226873030
```
Watch out for integer overflow.

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