# PROBLEM LINK:

Practice

Contest: Division 1

Contest: Division 2

Contest: Division 3

Contest: Division 4

* Author:* raysh07

*apoorv_me*

**Tester:***iceknight1093*

**Editorialist:**# DIFFICULTY:

TBD

# PREREQUISITES:

None

# PROBLEM:

Teams from N colleges attend jadavpur University’s fest. The i-th team has A_i members.

Each room can accommodate two people, and students from different colleges will not stay in the same room.

What’s the minimum number of moves needed to accommodate everyone?

# EXPLANATION:

Since people from different colleges won’t share a room, we can find the number of rooms needed for each college separately and add them up to get the overall answer.

So, look at a college team with A_i members. Since each room can hold two people:

- If A_i is even, \frac{A_i}{2} rooms will suffice.
- If A_i is odd, \frac{A_i + 1}{2} rooms are needed (one extra room for the extra person).

This can be written as \left\lceil \frac{A_i}{2} \right\rceil, where \left\lceil \ \ \right\rceil denotes the ceiling function.

The overall answer is thus

# TIME COMPLEXITY:

\mathcal{O}(N) per testcase.

# CODE:

## Editorialist's code (Python)

```
for _ in range(int(input())):
n = int(input())
print(sum([(x+1)//2 for x in map(int, input().split())]))
```