diff --git a/Py3Scripts/AwesomeMath.ipynb b/Py3Scripts/AwesomeMath.ipynb
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+++ b/Py3Scripts/AwesomeMath.ipynb
@@ -1 +1 @@
-{"cells":[{"cell_type":"code","execution_count":1,"metadata":{},"outputs":[],"source":"# 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？\n# 经典三元一次方程求解，设各有x，y，z只\n\n# 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\n# import time\n# start = time.perf_counter_ns()    # 用自带time函数统计运行时长\nfor x in range(0, 101, 5):  # 公鸡花费x元在0-100范围包括100，步长为5\n    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\n        z = 100 - x - y  # 小鸡花费z元为100元减去x和y\n        if x / 5 + y / 3 + z * 3 == 100:\n            print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x / 5, y / 3, z * 3))\n            # pass\n# end = time.perf_counter_ns()\n# time1 = end - start\n# print(\"解法一花费时间：\", time1)"},{"cell_type":"code","execution_count":2,"metadata":{},"outputs":[],"source":"# 解法二：解法和解法一类似\n# 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\n# 少，100元不再是固定的。假设花费钱数依次为x、y、z元\nfor x in range(0, 101, 5):  # 公鸡花费x元在0-100范围包括100，步长为5\n    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\n        for z in range(0, 101 - x - y):\n            if x / 5 + y / 3 + z * 3 == 100 and x + y + z == 100:  # 花费和鸡数都是100\n                print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x / 5, y / 3, z * 3))"},{"cell_type":"code","execution_count":3,"metadata":{},"outputs":[],"source":"# 解法三：枚举法\n# 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\n# 之间不包括100\nfor x in range(0, 20):\n    for y in range(0, 33):\n        z = 100 - x - y  # 小鸡个数z等于100只减去公鸡x只加母鸡y只\n        if 5 * x + 3 * y + z / 3 == 100:  # 钱数相加等于100元\n            print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x, y, z))"}],"nbformat":4,"nbformat_minor":2,"metadata":{"language_info":{"name":"python","codemirror_mode":{"name":"ipython","version":3}},"orig_nbformat":2,"file_extension":".py","mimetype":"text/x-python","name":"python","npconvert_exporter":"python","pygments_lexer":"ipython3","version":3}}
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+{"cells":[{"cell_type":"code","execution_count":5,"metadata":{},"outputs":[],"source":"# 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？\n# 经典三元一次方程求解，设各有x，y，z只\n\n# 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\n# import time\n# start = time.perf_counter_ns()    # 用自带time函数统计运行时长\nfor x in range(0, 101, 5):  # 公鸡花费x元在0-100范围包括100，步长为5\n    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\n        z = 100 - x - y  # 小鸡花费z元为100元减去x和y\n        if x / 5 + y / 3 + z * 3 == 100:\n            print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x / 5, y / 3, z * 3))\n            # pass\n# end = time.perf_counter_ns()\n# time1 = end - start\n# print(\"解法一花费时间：\", time1)\n\n# 解法二：解法和解法一类似\n# 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\n# 少，100元不再是固定的。假设花费钱数依次为x、y、z元\nfor x in range(0, 101, 5):  # 公鸡花费x元在0-100范围包括100，步长为5\n    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\n        for z in range(0, 101 - x - y):\n            if x / 5 + y / 3 + z * 3 == 100 and x + y + z == 100:  # 花费和鸡数都是100\n                print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x / 5, y / 3, z * 3))\n\n# 解法三：枚举法\n# 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\n# 之间不包括100\nfor x in range(0, 20):\n    for y in range(0, 33):\n        z = 100 - x - y  # 小鸡个数z等于100只减去公鸡x只加母鸡y只\n        if 5 * x + 3 * y + z / 3 == 100:  # 钱数相加等于100元\n            print(\"公鸡：%d只，母鸡：%d只，小鸡：%d只\" % (x, y, z))"},{"cell_type":"code","execution_count":2,"metadata":{},"outputs":[],"source":"# 经典斐波那契数列\n# 定义:https://wikimedia.org/api/rest_v1/media/math/render/svg/c374ba08c140de90c6cbb4c9b9fcd26e3f99ef56\n# 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\n\n# 方法一：使用递归\ndef fib1(n):\n    if n<0:\n        print(\"Incorrect input\")\n    elif n==1:\n        return 0    # 第一个斐波那契数是0\n    elif n==2:\n        return 1    # 第二斐波那契数是1\n    else:\n        return fib1(n-1)+fib1(n-2)\n\nprint(fib1(2))\n\n\n# 方法二：使用动态编程\nFibArray = [0, 1]\n\n\ndef fib2(n):\n    if n < 0:\n        print(\"Incorrect input\")\n    elif n <= len(FibArray):\n        return FibArray[n - 1]\n    else:\n        temp_fib = fib2(n - 1) + fib2(n - 2)\n        FibArray.append(temp_fib)\n        return temp_fib\n\n# 方法三：空间优化\ndef fibonacci(n):\n    a = 0\n    b = 1\n    if n < 0:\n        print(\"Incorrect input\")\n    elif n == 0:\n        return a\n    elif n == 1:\n        return b\n    else:\n        for i in range(2,n):\n            c = a + b\n            a = b\n            b = c\n        return b"},{"cell_type":"code","execution_count":3,"metadata":{},"outputs":[],"source":"# 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\n# 找出1-1000之间的水仙花数\n# 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\n\n# 方法一：解答四个数组成不重复三位数(暂未想到更优方法)\nfor x in range(1, 5):\n    for y in range(1, 5):\n        for z in range(1, 5):\n            if (x != y) and (x != z) and (z != y):\n                print(x, y, z)"},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":"# 计算pi小数点任意位数\nfrom __future__ import division\nimport math\nfrom time import time\ntime1 = time()\nnumber = int(input('输入计算的位数：'))\nnumber1 = number + 10  # 多计算十位方式尾数取舍影响\nb = 10 ** number1\n# 求含4/5的首项\nx1 = b * 4 // 5\n# 求含1/239的首项\nx2 = b // -239\n\n# 求第一大项\nhe = x1 + x2\n# 设置下面循环的终点，即共计算n项\nnumber *= 2\n\n# 循环初值=3，末值2n,步长=2\nfor i in range(3, number, 2):\n    # 求每个含1/5的项及符号\n    x1 //= -25\n    # 求每个含1/239的项及符号\n    x2 //= -57121\n    # 求两项之和\n    x = (x1 + x2) // i\n    # 求总和\n    he += x\n\n# 求出π\npi = he * 4\n# 舍掉后十位\npi //= 10 ** 10\n\n# 输出圆周率π的值\npi_string = str(pi)\nresult = pi_string[0] + str('.') + pi_string[1:len(pi_string)]\nprint(result)\n\ntime2 = time()\n\nprint(u'耗时：' + str(time2 - time1) + 's')\n\n\n# 使用chudnovsky算法计算\n# 参考链接：https://www.craig-wood.com/nick/articles/pi-chudnovsky/\n\n\"\"\"\nPython3 program to calculate Pi using python long integers, BINARY\nsplitting and the Chudnovsky algorithm\n\n\"\"\"\n\nimport math\nfrom gmpy2 import mpz\nfrom time import time\n\ndef pi_chudnovsky_bs(digits):\n    \"\"\"\n    Compute int(pi * 10**digits)\n\n    This is done using Chudnovsky's series with BINARY splitting\n    \"\"\"\n    C = 640320\n    C3_OVER_24 = C**3 // 24\n    def bs(a, b):\n        \"\"\"\n        Computes the terms for binary splitting the Chudnovsky infinite series\n\n        a(a) = +/- (13591409 + 545140134*a)\n        p(a) = (6*a-5)*(2*a-1)*(6*a-1)\n        b(a) = 1\n        q(a) = a*a*a*C3_OVER_24\n\n        returns P(a,b), Q(a,b) and T(a,b)\n        \"\"\"\n        if b - a == 1:\n            # Directly compute P(a,a+1), Q(a,a+1) and T(a,a+1)\n            if a == 0:\n                Pab = Qab = mpz(1)\n            else:\n                Pab = mpz((6*a-5)*(2*a-1)*(6*a-1))\n                Qab = mpz(a*a*a*C3_OVER_24)\n            Tab = Pab * (13591409 + 545140134*a) # a(a) * p(a)\n            if a & 1:\n                Tab = -Tab\n        else:\n            # Recursively compute P(a,b), Q(a,b) and T(a,b)\n            # m is the midpoint of a and b\n            m = (a + b) // 2\n            # Recursively calculate P(a,m), Q(a,m) and T(a,m)\n            Pam, Qam, Tam = bs(a, m)\n            # Recursively calculate P(m,b), Q(m,b) and T(m,b)\n            Pmb, Qmb, Tmb = bs(m, b)\n            # Now combine\n            Pab = Pam * Pmb\n            Qab = Qam * Qmb\n            Tab = Qmb * Tam + Pam * Tmb\n        return Pab, Qab, Tab\n    # how many terms to compute\n    DIGITS_PER_TERM = math.log10(C3_OVER_24/6/2/6)\n    N = int(digits/DIGITS_PER_TERM + 1)\n    # Calclate P(0,N) and Q(0,N)\n    P, Q, T = bs(0, N)\n    one_squared = mpz(10)**(2*digits)\n    sqrtC = (10005*one_squared).sqrt()\n    return (Q*426880*sqrtC) // T\n\n# The last 5 digits or pi for various numbers of digits\ncheck_digits = {\n    100 : 70679,\n    1000 :  1989,\n    10000 : 75678,\n    100000 : 24646,\n    1000000 : 58151,\n    10000000 : 55897,\n}\n\nif __name__ == \"__main__\":\n    digits = 100\n    pi = pi_chudnovsky_bs(digits)\n    print(pi)\n    #raise SystemExit\n    for log10_digits in range(1,9):\n        digits = 10**log10_digits\n        start =time()\n        pi = pi_chudnovsky_bs(digits)\n        print(\"chudnovsky_gmpy_mpz_bs: digits\",digits,\"time\",time()-start)\n        if digits in check_digits:\n            last_five_digits = pi % 100000\n            if check_digits[digits] == last_five_digits:\n                print(\"Last 5 digits %05d OK\" % last_five_digits)\n            else:\n                print(\"Last 5 digits %05d wrong should be %05d\" % (last_five_digits, check_digits[digits]))\n"},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":""}],"nbformat":4,"nbformat_minor":2,"metadata":{"language_info":{"name":"python","codemirror_mode":{"name":"ipython","version":3}},"orig_nbformat":2,"file_extension":".py","mimetype":"text/x-python","name":"python","npconvert_exporter":"python","pygments_lexer":"ipython3","version":3}}
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