Overview

Starting from an initial state with a single colony of size \(1\), the operation “every colony generates a new colony of twice its own size” is repeated \(K\) times. The problem asks how many distinct colony sizes exist at the end.

Analysis

The key to this problem is organizing the reproduction rules and observing how the “set of sizes” changes at each step.

1. Initial state (0 reproductions)

   • Colony sizes present: \(\{1\}\)
   • Number of distinct sizes: \(1\) (only \(2^0\))

2. After 1st reproduction

   • Original sizes: \(\{1\}\)
   • Newly created sizes: \(1 \times 2 = 2\)
   • All sizes: \(\{1, 2\}\)
   • Number of distinct sizes: \(2\) (\(2^0, 2^1\))

3. After 2nd reproduction

   • Original sizes: \(\{1, 2\}\)
   • Newly created sizes: \(1 \times 2 = 2, \ 2 \times 2 = 4\)
   • All sizes: \(\{1, 2, 2, 4\}\)
   • Distinct (removing duplicates): \(\{1, 2, 4\}\)
   • Number of distinct sizes: \(3\) (\(2^0, 2^1, 2^2\))

4. After 3rd reproduction

   • Original sizes: \(\{1, 2, 2, 4\}\)
   • Newly created sizes: \(2, 4, 4, 8\)
   • All sizes: \(\{1, 2, 2, 4, 2, 4, 4, 8\}\)
   • Distinct (removing duplicates): \(\{1, 2, 4, 8\}\)
   • Number of distinct sizes: \(4\) (\(2^0, 2^1, 2^2, 2^3\))

Key Observation

As reproductions are repeated, the new size “current maximum size \(\times 2\)” is always added. After the \(K\)-th reproduction, the distinct colony sizes are \(\{2^0, 2^1, 2^2, \dots, 2^K\}\). The number of elements in this set is \(K + 1\).

Why simulation doesn’t work

Looking at the constraints, \(K \leq 10^{18}\), which is an extremely large value. If you actually build a list of colonies or simulate the reproductions one by one, you will get a time limit exceeded (TLE) or run out of memory. However, from the analysis above, we can derive the rule that “the answer is always \(K + 1\),” so we can produce the answer without any complex computation.

Algorithm

1. Read input \(K\).
2. Compute and output \(K + 1\).

Complexity

• Time complexity: \(O(1)\)
  • Regardless of the value of \(K\), the answer is obtained with a single addition.
• Space complexity: \(O(1)\)
  • No special data structures are needed.

Implementation Notes

• Handling large numbers: Since \(K\) can be as large as \(10^{18}\), depending on the programming language, you need to use a 64-bit integer type (in Python, the standard int suffices).

• Reading input: Since \(K\) can be very large, it is common practice to read it as a string and then convert it to a numeric value.

  Source Code

import sys

def solve():
    # 標準入力から K を読み込む
    # 制約: 1 <= K <= 10^18
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    k = int(input_data[0])
    
    # 実験の推移を考える：
    # 開始時: サイズ {1} のコロニーが存在する（種類数: 1）
    # 1回目の増殖: {1} からサイズ 2 のコロニーが生まれ、元の 1 も残る -> {1, 2} （種類数: 2）
    # 2回目の増殖: {1, 2} のそれぞれから 2倍のサイズ (2, 4) が生まれ、元も残る -> {1, 2, 2, 4} （種類数: 3）
    # 3回目の増殖: {1, 2, 2, 4} から (2, 4, 4, 8) が生まれ、元も残る -> {1, 2, 2, 4, 2, 4, 4, 8} （種類数: 4）
    # 
    # このように、K 回の増殖後には {2^0, 2^1, 2^2, ..., 2^K} という K+1 種類のサイズが存在することになる。
    # したがって、求める種類数は K + 1 である。
    
    print(k + 1)

if __name__ == '__main__':
    solve()

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