feat(0053): 添加组合数超百万计数解法

solutions/0053.Combinatoric/euler_53.py

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+"""
+There are exactly ten ways of selecting three from five, 12345:
+
+123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
+
+In combinatorics, we use the notation, (5 3) = 10 .
+
+In general, (n r) = n! / (r! * (n - r)!) , where r ≤ n , n!=n×(n-1)×...×3×2×1 , and 0!=1 .
+
+It is not until n=23 , that a value exceeds one-million: (23 10) = 1144066 .
+
+How many, not necessarily distinct, values of (n r) for 1 ≤ n ≤ 100 , are greater than one-million?
+"""
+
+import time
+from functools import lru_cache
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        elapsed_time = end_time - start_time
+        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def comfun(n, r):
+    return factorial(n) // (factorial(r) * factorial(n - r))
+
+
+@lru_cache(maxsize=None)
+def factorial(n: int, s: int = 1):
+    if n < s:
+        return 1
+    elif n == s:
+        if s < 1:
+            return 1
+        return s
+    else:
+        return n * factorial(n - 1)
+
+
+@timer
+def main():
+    num = 0
+    for i in range(1, 101):
+        for j in range(1, i + 1):
+            if comfun(i, j) > 1000000:
+                num += 1
+    print(num)
+
+
+if __name__ == "__main__":
+    main()