UPD:Minor mistakes in grammar and expression fixed.

Disclaimer: This is not an official editorial.

If you have better or easy-to-understand solutions and thoughts, feel free to share your idea. Also welcome to point out mistakes so I can fix it.

### 160A - Близнецы

It's obvious that you should take the most valueable coins. so sort values in non-decreasing order, then take coins from the most valueable to the least, until you get **strictly** more than half of total value.

Time complexity depends on the sorting algorithm you use. O(n^2) is also acceptable, but if you use bogosort which runs in O(n!)...

### 160B - Несчастливый билет

Deal with the situation that "first half is strictly less than second half" first. the other one can be solved accordingly.

You can use greedy here: sort digits in first and second half seperately. then if the i-th digit in first half is always less than i-th in second half for 1<=i<=n, answer is YES.

Time complexity is as same as problem A. Count sort may run faster in O(n+10), but it's not necessary