Balls and boxes#9890

Balls and boxes (Editorial)

Let's consider the first ball. It can be placed in one of the $k$ available boxes. Thus, there are $k$ options for placing the first ball. Similarly, each subsequent ball can also be placed in one of the $k$ boxes (there are $k$ options for each ball).

Therefore, the total number of ways to distribute all the balls into the boxes is the product $k \cdot k \cdot ... \cdot k = k^{n}$ k \cdot k \cdot ... \cdot k = k^n.

Example

Let's say we have two balls numbered $1$ and $2$ , as well as two boxes numbered $1$ and $2$ . In this case, the balls can be distributed into the boxes in the following four ways:

Algorithm implementation

Read the input data.

scanf("%lld %lld", &n, &k);

Compute $r es = k^{n}$ res = k^n.

res = 1;
for (i = 1; i <= n; i++)
  res = res * k;

Print the answer.

printf("%lld\n", res);