Candies for Nephews

This problem is just divisibility by $3$ wearing a tiny fake mustache.

We have $n$ candies, and Monocarp has exactly three nephews. He wants each nephew to get the same number of candies, and he must give away all candies he already has. If he buys $x$ extra candies, then the final total is:

$$n + x$$

For this total to be split equally among three nephews, it must be divisible by $3$:

$$(n+x) \bmod 3 = 0$$

So the task is: find the smallest nonnegative $x$ that makes $n+x$ a multiple of $3$.

Checking the remainder

Only the remainder of $n$ after division by $3$ matters.

There are three cases:

• If $n \bmod 3 = 0$, then $n$ is already divisible by $3$. Buy $0$ candies.
• If $n \bmod 3 = 1$, then $n$ is one candy past a multiple of $3$. The next multiple is reached by adding $2$. Buy $2$ candies.
• If $n \bmod 3 = 2$, then $n$ is two candies past a multiple of $3$. The next multiple is reached by adding $1$. Buy $1$ candy.

Example: $n=7$.

$$7 \bmod 3 = 1$$

So we buy $2$, giving a total of $9$, and $9$ splits into $3,3,3$.

Example: $n=24$.

$$24 \bmod 3 = 0$$

Already good. Buy nothing.

Compact formula

The casework can be written as:

$$\text{answer} = (3 - (n \bmod 3)) \bmod 3$$

Why the extra modulo at the end? Because if $n \bmod 3 = 0$, then $3 - 0 = 3$, but buying $3$ candies is not minimal. Buying $0$ is. The final modulo converts that $3$ back to $0$.

So for every test case, compute the remainder and print the smallest number needed to reach the next multiple of $3$.

The constraints are tiny, but even if they were huge, this would still be $O(1)$ per test case. No loops over candies, no simulation, no nonsense. Just modulo and done.