问:
【基础题】:求 1+2!+3!+...+20! 的和
【提高题】:两个乒乓球队进行比赛,各出三人。甲队为 a,b,c 三人,乙队为 x,y,z 三人。已抽签决定比赛名单。 有人向队员打听比赛的名单。 a 说他不和 x 比, c 说他不和 x,z 比,请找出三队赛手的名单
答:
方法1:
def factorial(n): if n == 0 or n == 1: return 1 else: return factorial(n - 1) * nmy_sum = sum(factorial(i) for i in range(1, 21))print(f"1! + 2! + 3! + 4! + 5! + ... + 20!={my_sum}")
方法2:
def recursion(n): #'定义递归函数实现求阶乘功能' if n == 1: return 1 else: return n*recursion(n-1)list_recursion = [] # 定义一个空的列表,将调用递归函数生成的阶乘值追加到列表for i in range(1,21): list_recursion.append(recursion(i)) # 将调用递归函数生成的阶乘值追加到列表print(sum(list_recursion))
方法3:
def recursion(n): #'定义递归函数实现求阶乘功能' if n == 1: return 1 else: return n*recursion(n-1)Sum = 0for i in range(1, 21): Sum += recursion(i)print(Sum)
方法4:
def factorial(num): if num == 1: return 1 else: return num * factorial(num - 1)def sum_factorial(): iteration = int(input("Input the max iteration:")) result = 0 for i in range(1, iteration+1): result += factorial(i) print("Alright, we catch the result:{}".format(result))if __name__ == '__main__': sum_factorial()
方法5:
def factorial(n): f = 1 for i in range(1, n+1): f = f * i return fnum = int(input("请输入一个数字:"))sum = 0if num < 0: print("抱歉,负数没有阶乘")elif num == 0: print("0的阶乘为1")else: for i in range(1, num+1): sum += factorial(i) print(sum)
方法6:
def factorial(number): start = 1 fact = 1 while start <= number: yield fact start += 1 fact *= startfact = factorial(20)sum = 0for i in fact: sum += iprint(sum)
提高题:两个乒乓球队进行比赛,各出三人。甲队为 a,b,c 三人,乙队为 x,y,z 三人。已抽签决定比赛名单。 有人向队员打听比赛的名单。 a 说他不和 x 比, c 说他不和 x,z 比,请找出三队赛手的名单。
from itertools import permutationsteam_a = ["a", "b", "c"]team_b = ["x", "y", "z"]possible_match_lists = []for possible_team_b_sequence in list(permutations(team_b)): possible_match_lists.append(dict(zip(team_a, possible_team_b_sequence)))final_match_lists = list(filter(lambda x: x["a"] != "x" and \ x["c"] != "x" and \ x["c"] != "z", possible_match_lists))for index, match_list in enumerate(final_match_lists): print(f"solution{index+1}") for competitiors in match_list.items(): print(" vs ".join(competitiors))