|
|
|
|
|
# 此文件用来记录经典或有趣的数学问题
|
|
|
|
|
|
# It's really fun to swim in the ocean of mathematics
|
|
|
|
|
|
|
|
|
|
|
|
# 百钱白鸡问题：1只公鸡5元，1只母鸡3元，3只小鸡1元，100元买100只鸡，问：公鸡母鸡小鸡各有多少？
|
|
|
|
|
|
# 经典三元一次方程求解，设各有x，y，z只
|
|
|
|
|
|
|
|
|
|
|
|
# 解法一：推断每种鸡花费依次轮询，运行时间最短，2019-7-24最优方案
|
|
|
|
|
|
# import time
|
|
|
|
|
|
# start = time.perf_counter_ns() # 用自带time函数统计运行时长
|
|
|
|
|
|
for x in range(0, 101, 5): # 公鸡花费x元在0-100范围包括100，步长为5
|
|
|
|
|
|
for y in range(0, 101 - x, 3): # 母鸡花费y元在0到100元减去公鸡花费钱数，步长为3
|
|
|
|
|
|
z = 100 - x - y # 小鸡花费z元为100元减去x和y
|
|
|
|
|
|
if x / 5 + y / 3 + z * 3 == 100:
|
|
|
|
|
|
print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3))
|
|
|
|
|
|
# pass
|
|
|
|
|
|
# end = time.perf_counter_ns()
|
|
|
|
|
|
# time1 = end - start
|
|
|
|
|
|
# print("解法一花费时间：", time1)
|
|
|
|
|
|
|
|
|
|
|
|
# 解法二：枚举法
|
|
|
|
|
|
# 解题思路：若只买公鸡最多20只，但要买100只，固公鸡在0-20之间不包括20;若只买母鸡则在0-33之间不包括33;若只买小鸡则在0-100
|
|
|
|
|
|
# 之间不包括100
|
|
|
|
|
|
for x in range(0, 20):
|
|
|
|
|
|
for y in range(0, 33):
|
|
|
|
|
|
z = 100 - x - y # 小鸡个数z等于100只减去公鸡x只加母鸡y只
|
|
|
|
|
|
if 5 * x + 3 * y + z / 3 == 100: # 钱数相加等于100元
|
|
|
|
|
|
print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x, y, z))
|
|
|
|
|
|
|
|
|
|
|
|
# 解法三：解法和解法一类似
|
|
|
|
|
|
# 解题思路：买一只公鸡花费5元，剩余95元(注意考虑到不买公鸡的情况)，再买一只母鸡花费3元剩余92元，依次轮询下去，钱数不断减
|
|
|
|
|
|
# 少，100元不再是固定的。假设花费钱数依次为x、y、z元
|
|
|
|
|
|
for x in range(0, 101, 5): # 公鸡花费x元在0-100范围包括100，步长为5
|
|
|
|
|
|
for y in range(0, 101 - x, 3): # 母鸡花费y元在0到100元减去公鸡花费钱数，步长为3
|
|
|
|
|
|
for z in range(0, 101 - x - y):
|
|
|
|
|
|
if x / 5 + y / 3 + z * 3 == 100 and x + y + z == 100: # 花费和鸡数都是100
|
|
|
|
|
|
print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3))
|
|
|
|
|
|
|
|
|
|
|
|
# 经典斐波那契数列
|
|
|
|
|
|
# 定义:https://wikimedia.org/api/rest_v1/media/math/render/svg/c374ba08c140de90c6cbb4c9b9fcd26e3f99ef56
|
|
|
|
|
|
# 用文字来说，就是斐波那契数列由0和1开始，之后的斐波那契系数就是由之前的两数相加而得出
|
|
|
|
|
|
# 方法一：使用递归
|
|
|
|
|
|
def fib1(n):
|
|
|
|
|
|
if n<0:
|
|
|
|
|
|
print("Incorrect input")
|
|
|
|
|
|
elif n==1:
|
|
|
|
|
|
return 0 # 第一个斐波那契数是0
|
|
|
|
|
|
elif n==2:
|
|
|
|
|
|
return 1 # 第二斐波那契数是1
|
|
|
|
|
|
else:
|
|
|
|
|
|
return fib1(n-1)+fib1(n-2)
|
|
|
|
|
|
|
|
|
|
|
|
print(fib1(2))
|
|
|
|
|
|
|
|
|
|
|
|
# 方法二：使用动态编程
|
|
|
|
|
|
FibArray = [0, 1]
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def fib2(n):
|
|
|
|
|
|
if n < 0:
|
|
|
|
|
|
print("Incorrect input")
|
|
|
|
|
|
elif n <= len(FibArray):
|
|
|
|
|
|
return FibArray[n - 1]
|
|
|
|
|
|
else:
|
|
|
|
|
|
temp_fib = fib2(n - 1) + fib2(n - 2)
|
|
|
|
|
|
FibArray.append(temp_fib)
|
|
|
|
|
|
return temp_fib
|
|
|
|
|
|
|
|
|
|
|
|
# 方法三：空间优化
|
|
|
|
|
|
|
|
|
|
|
|
def fibonacci(n):
|
|
|
|
|
|
a = 0
|
|
|
|
|
|
b = 1
|
|
|
|
|
|
if n < 0:
|
|
|
|
|
|
print("Incorrect input")
|
|
|
|
|
|
elif n == 0:
|
|
|
|
|
|
return a
|
|
|
|
|
|
elif n == 1:
|
|
|
|
|
|
return b
|
|
|
|
|
|
else:
|
|
|
|
|
|
for i in range(2,n):
|
|
|
|
|
|
c = a + b
|
|
|
|
|
|
a = b
|
|
|
|
|
|
b = c
|
|
|
|
|
|
return b
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
# 水仙花数：水仙花数即此数字是各位立方和等于这个数本身的数。例：153 = 1**3 + 5**3 + 3**3
|
|
|
|
|
|
# 找出1-1000之间的水仙花数
|
|
|
|
|
|
|
|
|
|
|
|
# 分别四个数字：1,2,3,4，组成不重复的三位数。问题扩展：对于给定数字或给定范围的数字，组成不重复的n位数
|
|
|
|
|
|
|
|
|
|
|
|
# 方法一：解答四个数组成不重复三位数(暂未想到更优方法)
|
|
|
|
|
|
for x in range(1, 5):
|
|
|
|
|
|
for y in range(1, 5):
|
|
|
|
|
|
for z in range(1, 5):
|
|
|
|
|
|
if (x != y) and (x != z) and (z != y):
|
|
|
|
|
|
print(x, y, z)
|
|
|
|
|
|
|
