Overview

This problem asks us to find the number of distinct sizes that exist after a bacterial colony has reproduced \(K\) times. By analyzing the reproduction pattern, we find that the answer can be obtained with the simple formula \(K + 1\).

Analysis

Simulating the Reproduction

First, let’s examine what colony sizes actually exist for small values of \(K\).

Initial state (\(K = 0\)) - Sizes: \(\{1\}\) (1 type)

After the 1st reproduction (\(K = 1\)) - The original colony (size 1) remains as is - From the size-1 colony, a new colony of size \(1 \times 2 = 2\) is born - Sizes: \(\{1, 2\}\) (2 types)

After the 2nd reproduction (\(K = 2\)) - From size 1 → size 2 is born - From size 2 → size 4 is born - Sizes: \(\{1, 2, 4\}\) (3 types)

After the 3rd reproduction (\(K = 3\)) - From size 1 → size 2 is born - From size 2 → size 4 is born - From size 4 → size 8 is born - Sizes: \(\{1, 2, 4, 8\}\) (4 types)

Key Observation

By observing this experiment, we can notice the following:

1. Sizes are always powers of 2: \(1 = 2^0\), \(2 = 2^1\), \(4 = 2^2\), \(8 = 2^3\), …
2. After \(K\) reproductions, the only sizes that exist are \(2^0, 2^1, 2^2, \ldots, 2^K\)
3. The number of distinct sizes is \(K + 1\)

The reason for this is that a colony of size \(2^i\) means “the initial size-1 colony that has been doubled \(i\) times.” With \(K\) reproductions, colonies that have undergone 0 through \(K\) doubling operations can exist, resulting in \(K + 1\) distinct types.

Problem with the Naive Approach

Since \(K\) can be as large as \(10^{18}\), it is impossible to actually perform the simulation. Running a loop \(K\) times would result in a time complexity of \(O(K)\), leading to TLE.

Solution

From the analysis above, we have established that the number of distinct colony sizes after \(K\) reproductions is always \(K + 1\). Therefore, we simply output \(K + 1\) directly without any loops.

Algorithm

1. Read input \(K\)
2. Output \(K + 1\)

Complexity

• Time complexity: \(O(1)\)
• Space complexity: \(O(1)\)

Implementation Notes

• Although \(K\) can be as large as \(10^{18}\), in Python there is no need to worry about integer overflow

• This is a classic mathematical insight problem where recognizing the essence of the problem eliminates the need for complex simulation

  Source Code

K = int(input())
print(K + 1)

---

This editorial was generated by claude4.5opus.