87 lines
2.3 KiB
Python
87 lines
2.3 KiB
Python
"""
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Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
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Triangle: T_n = n(n+1)/2
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Pentagonal: P_n = n(3n-1)/2
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Hexagonal: H_n = n(2n-1)
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It can be verified that T_285 = P_165 = H_143 = 40755.
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Find the next triangle number that is also pentagonal and hexagonal.
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"""
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import time
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def timer(func):
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def wrapper(*args, **kwargs):
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start_time = time.time()
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result = func(*args, **kwargs)
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end_time = time.time()
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print(f"{func.__name__} taken: {end_time - start_time:.6f} seconds")
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return result
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return wrapper
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def is_pentagonal(x):
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"""检查一个数是否为五边形数"""
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# 解方程 n(3n-1)/2 = x => 3n² - n - 2x = 0
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# 判别式 Δ = 1 + 24x 必须是完全平方数
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delta = 1 + 24 * x
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sqrt_delta = int(delta**0.5)
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if sqrt_delta * sqrt_delta != delta:
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return False
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# 并且 (1 + sqrt(Δ)) ≡ 0 mod 6
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return (1 + sqrt_delta) % 6 == 0
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def is_hexagonal(x):
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"""检查一个数是否为六边形数"""
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# 解方程 n(2n-1) = x => 2n² - n - x = 0
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# 判别式 Δ = 1 + 8x 必须是完全平方数
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delta = 1 + 8 * x
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sqrt_delta = int(delta**0.5)
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if sqrt_delta * sqrt_delta != delta:
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return False
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# 并且 (1 + sqrt(Δ)) ≡ 0 mod 4
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return (1 + sqrt_delta) % 4 == 0
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@timer
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def main() -> None:
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n = 286
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while True:
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triangle = n * (n + 1) // 2
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# 直接使用已有的验证函数，避免重复计算
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if is_pentagonal(triangle) and is_hexagonal(triangle):
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# 找到后计算对应的索引
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pent_index = (1 + int((1 + 24 * triangle) ** 0.5)) // 6
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hex_index = (1 + int((1 + 8 * triangle) ** 0.5)) // 4
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print(f"Answer: T({n})=P({pent_index})=H({hex_index})={triangle}")
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break
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n += 1
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@timer
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def main_quick_math(nth: int = 2) -> None:
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u = 5
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v = 3
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sn = 0
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while True:
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u, v = (2 * u + 3 * v), (u + 2 * v)
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if u % 6 == 5 and v % 4 == 3:
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k = (u + 1) // 6
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j = (v + 1) // 4
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n = 2 * j - 1
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sn += 1
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if sn == nth:
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print(f"Answer: T({n})=P({k})=H({j})={n * (n + 1) // 2}")
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break
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if __name__ == "__main__":
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main()
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main_quick_math()
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