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# 此文件用来记录经典或有趣的数学问题
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# It's really fun to swim in the ocean of mathematics
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# 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？
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# 经典三元一次方程求解，设各有x，y，z只
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# 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
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# import time
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# start = time.perf_counter_ns() # 用自带time函数统计运行时长
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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
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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
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z = 100 - x - y # 小鸡花费z元为100元减去x和y
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if x / 5 + y / 3 + z * 3 == 100:
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print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3))
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# pass
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# end = time.perf_counter_ns()
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# time1 = end - start
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# print("解法一花费时间：", time1)
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# 解法二：枚举法
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# 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
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# 之间不包括100
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for x in range(0, 20):
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for y in range(0, 33):
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z = 100 - x - y # 小鸡个数z等于100只减去公鸡x只加母鸡y只
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if 5 * x + 3 * y + z / 3 == 100: # 钱数相加等于100元
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print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x, y, z))
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# 解法三：解法和解法一类似
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# 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
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# 少，100元不再是固定的。假设花费钱数依次为x、y、z元
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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
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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
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for z in range(0, 101 - x - y):
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if x / 5 + y / 3 + z * 3 == 100 and x + y + z == 100: # 花费和鸡数都是100
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print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3))
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# 经典斐波那契数列
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# 定义:https://wikimedia.org/api/rest_v1/media/math/render/svg/c374ba08c140de90c6cbb4c9b9fcd26e3f99ef56
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# 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
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# 方法一：使用递归
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def fib1(n):
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if n<0:
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print("Incorrect input")
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elif n==1:
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return 0 # 第一个斐波那契数是0
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elif n==2:
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return 1 # 第二斐波那契数是1
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else:
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return fib1(n-1)+fib1(n-2)
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print(fib1(2))
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# 方法二：使用动态编程
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FibArray = [0, 1]
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def fib2(n):
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if n < 0:
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print("Incorrect input")
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elif n <= len(FibArray):
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return FibArray[n - 1]
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else:
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temp_fib = fib2(n - 1) + fib2(n - 2)
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FibArray.append(temp_fib)
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return temp_fib
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# 方法三：空间优化
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def fibonacci(n):
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a = 0
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b = 1
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if n < 0:
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print("Incorrect input")
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elif n == 0:
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return a
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elif n == 1:
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return b
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else:
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for i in range(2,n):
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c = a + b
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a = b
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b = c
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return b
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# 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
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# 找出1-1000之间的水仙花数
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# 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
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# 方法一：解答四个数组成不重复三位数(暂未想到更优方法)
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for x in range(1, 5):
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for y in range(1, 5):
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for z in range(1, 5):
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if (x != y) and (x != z) and (z != y):
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print(x, y, z)
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# 计算pi小数点任意位数
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from __future__ import division
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import math
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from time import time
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time1 = time()
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number = int(input('输入计算的位数：'))
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number1 = number + 10 # 多计算十位方式尾数取舍影响
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b = 10 ** number1
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# 求含4/5的首项
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x1 = b * 4 // 5
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# 求含1/239的首项
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x2 = b // -239
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# 求第一大项
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he = x1 + x2
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# 设置下面循环的终点，即共计算n项
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number *= 2
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# 循环初值=3，末值2n,步长=2
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for i in range(3, number, 2):
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# 求每个含1/5的项及符号
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x1 //= -25
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# 求每个含1/239的项及符号
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x2 //= -57121
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# 求两项之和
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x = (x1 + x2) // i
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# 求总和
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he += x
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# 求出π
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pi = he * 4
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# 舍掉后十位
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pi //= 10 ** 10
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# 输出圆周率π的值
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pi_string = str(pi)
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result = pi_string[0] + str('.') + pi_string[1:len(pi_string)]
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print(result)
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time2 = time()
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print(u'耗时：' + str(time2 - time1) + 's')
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# 使用chudnovsky算法计算
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# 理解链接：https://www.craig-wood.com/nick/articles/pi-chudnovsky/
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"""
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Python3 program to calculate Pi using python long integers, BINARY
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splitting and the Chudnovsky algorithm
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"""
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import math
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from gmpy2 import mpz
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from time import time
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def pi_chudnovsky_bs(digits):
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"""
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Compute int(pi * 10**digits)
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This is done using Chudnovsky's series with BINARY splitting
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"""
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C = 640320
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C3_OVER_24 = C**3 // 24
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def bs(a, b):
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"""
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Computes the terms for binary splitting the Chudnovsky infinite series
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a(a) = +/- (13591409 + 545140134*a)
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p(a) = (6*a-5)*(2*a-1)*(6*a-1)
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b(a) = 1
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q(a) = a*a*a*C3_OVER_24
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returns P(a,b), Q(a,b) and T(a,b)
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"""
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if b - a == 1:
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# Directly compute P(a,a+1), Q(a,a+1) and T(a,a+1)
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if a == 0:
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Pab = Qab = mpz(1)
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else:
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Pab = mpz((6*a-5)*(2*a-1)*(6*a-1))
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Qab = mpz(a*a*a*C3_OVER_24)
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Tab = Pab * (13591409 + 545140134*a) # a(a) * p(a)
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if a & 1:
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Tab = -Tab
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else:
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# Recursively compute P(a,b), Q(a,b) and T(a,b)
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# m is the midpoint of a and b
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m = (a + b) // 2
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# Recursively calculate P(a,m), Q(a,m) and T(a,m)
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Pam, Qam, Tam = bs(a, m)
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# Recursively calculate P(m,b), Q(m,b) and T(m,b)
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Pmb, Qmb, Tmb = bs(m, b)
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# Now combine
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Pab = Pam * Pmb
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Qab = Qam * Qmb
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Tab = Qmb * Tam + Pam * Tmb
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return Pab, Qab, Tab
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# how many terms to compute
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DIGITS_PER_TERM = math.log10(C3_OVER_24/6/2/6)
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N = int(digits/DIGITS_PER_TERM + 1)
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# Calclate P(0,N) and Q(0,N)
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P, Q, T = bs(0, N)
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one_squared = mpz(10)**(2*digits)
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sqrtC = (10005*one_squared).sqrt()
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return (Q*426880*sqrtC) // T
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# The last 5 digits or pi for various numbers of digits
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check_digits = {
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100 : 70679,
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1000 : 1989,
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10000 : 75678,
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100000 : 24646,
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1000000 : 58151,
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10000000 : 55897,
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}
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if __name__ == "__main__":
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digits = 100
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pi = pi_chudnovsky_bs(digits)
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print(pi)
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#raise SystemExit
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for log10_digits in range(1,9):
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digits = 10**log10_digits
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start =time()
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pi = pi_chudnovsky_bs(digits)
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print("chudnovsky_gmpy_mpz_bs: digits",digits,"time",time()-start)
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if digits in check_digits:
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last_five_digits = pi % 100000
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if check_digits[digits] == last_five_digits:
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print("Last 5 digits %05d OK" % last_five_digits)
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else:
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print("Last 5 digits %05d wrong should be %05d" % (last_five_digits, check_digits[digits]))
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root
5 years ago