Limits
1s, 512 MB
The Koch snowflake can be constructed by starting with an equilateral triangle, then recursively altering each line segment as follows:
- divide the line segment into three segments of equal length.
- draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
- remove the line segment that is the base of the triangle from step 2.
Below is a figure of Koch snowflake of order 1, 2, 3 and 4.
Given the order N, of the Koch Snowflake, you need to find the number of peak vertices and edges of the koch snowflake.
Input
First line will contain T (T≤10000), the number of test cases. Each of the T lines will contain one integer N (0<N≤1018).
Output
For each case, print the number of vertices and edges for Koch Snowflake of order N modulo 1,000,000,007.
Sample
Input | Output |
---|
2
1
2
| Case 1: 3 3
Case 2: 6 12
|
Factors
| CPU | Memory | Source |
---|
Bash 5.2 | 1× | 1× | 1× |
Brainf*ck | 1× | 1× | 1× |
C# Mono 6.0 | 1× | 1× | 1× |
C++17 GCC 13.2 | 1× | 1× | 1× |
C++20 Clang 16.0 | 1× | 1× | 1× |
C++20 GCC 13.2 | 1× | 1× | 1× |
C++23 GCC 13.2 | 1× | 1× | 1× |
C11 GCC 13.2 | 1× | 1× | 1× |
C17 GCC 13.2 | 1× | 1× | 1× |
C23 GCC 13.2 | 1× | 1× | 1× |
Common Lisp SBCL 2.0 | 1× | 1× | 1× |
D8 11.8 | 1× | 1× | 1× |
Erlang 22.3 | 1× | 1× | 1× |
Free Pascal 3.0 | 1× | 1× | 1× |
Go 1.22 | 1× | 1× | 1× |
Grep 3.7 | 1× | 1× | 1× |
Haskell 8.6 | 1× | 1× | 1× |
Java 1.8 | 1× | 1× | 1× |
Kotlin 1.1 | 1× | 1× | 1× |
Lua 5.4 | 1× | 1× | 1× |
Node.js 10.16 | 1× | 1× | 1× |
Perl 5.30 | 1× | 1× | 1× |
PHP 8.3 | 1× | 1× | 1× |
PyPy 7.1 (3.6) | 1× | 1× | 1× |
Python 3.12 | 1× | 1× | 1× |
Ruby 3.2 | 1× | 1× | 1× |
Rust 1.57 | 1× | 1× | 1× |
Swift 5.3 | 1× | 1× | 1× |
Whitespace | 1× | 1× | 1× |