"""
The nth term of the sequence of triangle numbers is given by, $t_n = \frac{n(n+1)}{2}$  ;
so the first ten triangle numbers are:
    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
By converting each letter in a word to a number corresponding to its alphabetical position
and adding these values we form a word value. For example, the word value for SKY is 19+11+25 = 55 = t_10 .
If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand
common English words, how many are triangle words?
"""

import time
from pathlib import Path


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.perf_counter()
        result = func(*args, **kwargs)
        end_time = time.perf_counter()
        print(f"Time taken by {func.__name__}: {end_time - start_time:.8f} seconds")
        return result

    return wrapper


def word_to_num(txt: str) -> int:
    """
    将单词转换为数字，每个字母对应其在字母表中的位置（A=1, B=2, ..., Z=26）

    参数:
        txt: 要转换的单词

    返回:
        单词的数字表示

    示例:
        >>> word_to_num("SKY")
        55
    """
    return sum(ord(c) - ord("A") + 1 for c in txt)


def is_triangle_number(n: int) -> bool:
    """
    判断一个数是否为三角形数

    参数:
        n: 要判断的数

    返回:
        True 如果 n 是三角形数，否则 False

    示例:
        >>> is_triangle_number(10)
        True
        >>> is_triangle_number(11)
        False
    """
    # 使用二次方程求解，如果结果是整数则为三角形数
    # 三角形数公式：n = k(k+1)/2
    # 解得 k = (-1 + sqrt(1 + 8n)) / 2
    k = (-1 + (1 + 8 * n) ** 0.5) / 2
    return k.is_integer()


@timer
def main() -> None:
    with open(Path(__file__).parent / "0042_words.txt", "r") as file:
        words = file.read().replace('"', "").split(",")

    triangle_words = [word for word in words if is_triangle_number(word_to_num(word))]
    print(f"Number of triangle words: {len(triangle_words)}")


if __name__ == "__main__":
    main()