feat(solutions/0057): 添加欧拉计划第57题连分数平方根解法

包含正算递推实现、约分优化及详细数学推导文档

solutions/0057.SquareRoot/euler_57.py

@@ -1,10 +1,26 @@
 '''
+It is possible to show that the square root of two can be expressed as an infinite continued fraction.
+sqrt{2}=1+frac{1}{2+frac{1}{2+frac{1}{2+...}}}
 
+By expanding this for the first four iterations, we get:
+begin{aligned}
+ & 1+frac{1}{2}=frac{3}{2}=1.5 \\
+ & 1+frac{1}{2+frac{1}{2}}=frac{7}{5}=1.4 \\
+ & 1+frac{1}{2+frac{1}{2+frac{1}{2}}}=frac{17}{12}=1.41666... \\
+ & 1+frac{1}{2+frac{1}{2+frac{1}{2+frac{1}{2}}}}=frac{41}{29}=1.41379 ldots
+end{aligned}
+
+The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, 
+is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
+
+In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?
 '''
 import time
 from functools import wraps
 from typing import Any, Callable, TypeVar
 
+from math import gcd
+
 F = TypeVar("F", bound=Callable[..., Any])
 
 
@@ -37,4 +53,38 @@ def benchmark(repeat: int = 1) -> Callable[[F], F]:
         wrapper.total_time = 0.0  # type: ignore[attr-defined]
         return wrapper  # type: ignore[return-value]
 
-    return decorator
+    return decorator
+
+
+def compute_square_root(n: int = 1000) -> list[tuple[int,int]] :
+    """
+    Compute the square root of 2 up to n expansions.
+    """
+    cf = [1] + [2 for _ in range(0, n)]
+    p2, p1 = 0, 1   # p_{-2}, p_{-1}
+    q2, q1 = 1, 0   # q_{-2}, q_{-1}
+    results = []
+    for a in cf:
+        p = a * p1 + p2
+        q = a * q1 + q2
+        p2, p1 = p1, p
+        q2, q1 = q1, q
+        if gc := gcd(p, q) != 1:
+            p //= gc
+            q //= gc
+        results.append((p, q))
+    return results
+
+@benchmark(20)
+def main() -> None:
+    results = compute_square_root(1000)
+    count = 0
+    for p, q in results:
+        if len(str(p)) > len(str(q)):
+            count += 1
+    print(f"{count} fractions contain a numerator with more digits than the denominator.")
+    print(f"sqrt{2} = {results[-1][0]/results[-1][1]:.15f}")
+
+
+if __name__ == "__main__":
+    main()