Ugh. Really weak. Problem E actually still has "expected possibilities" to solve it, but I given up.

## E. Travelling Salesman Problem

### Solution

Not solve this problem by myself. Looked the solution above.

### Code

## D. 3-Coloring

Interactive problem is interesting. It's a game problem but more like a construction problem. Not very hard. But my solution is a little hard to implementation. Not sure if there is other solutions.

## C. Balance the Bits

### Solution

Classic bracket pairing problem. Calculate prefix sum from left to right. So if found '(' then +1, otherwise -1. Of course in the whole process, the total sum should be zero and non of prefix less than 0.

Now think about $s_i=0$. That means in $a$ and $b$, one of them +1 another one -1. We just +1 for the one which prefix sum is less than another.

Then think about $s_i=1$. This one will affect the prefix sum for both $a$ and $b$. Because $s_i=0$ will need -1. Also because two 0 in $s$ will not impact the prefix sum. So for more safety, first half of $1$, we always chose '('. And other half always ')'.

### Code

Get WA once because of a small mistake. I forgot check if anything left after scanned $s$.

## B. Flip the Bits

### Solution

Do the operation from back to the front. Just a implementation problem.

### Code

## A. Déjà Vu

### Solution

Implementation. Try insert to the start and end position. If both of them are palindrome, then return no. Otherwise, print the one which is not palindrome.