{"cells":[{"cell_type":"code","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\n","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\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\n","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\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\n","for 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","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\n","def 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","\n","print(fib1(2))\n","\n","\n","# 方法二：使用动态编程\n","FibArray = [0, 1]\n","\n","\n","def 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\n","def 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","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)\n","for 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","metadata":{},"outputs":[],"source":["# 计算pi小数点任意位数\n","from __future__ import division\n","import math\n","from time import time\n","time1 = time()\n","number = int(input('输入计算的位数：'))\n","number1 = number + 10  # 多计算十位方式尾数取舍影响\n","b = 10 ** number1\n","# 求含4/5的首项\n","x1 = b * 4 // 5\n","# 求含1/239的首项\n","x2 = b // -239\n","\n","# 求第一大项\n","he = x1 + x2\n","# 设置下面循环的终点，即共计算n项\n","number *= 2\n","\n","# 循环初值=3，末值2n,步长=2\n","for 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\n","pi = he * 4\n","# 舍掉后十位\n","pi //= 10 ** 10\n","\n","# 输出圆周率π的值\n","pi_string = str(pi)\n","result = pi_string[0] + str('.') + pi_string[1:len(pi_string)]\n","print(result)\n","\n","time2 = time()\n","\n","print(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","\"\"\"\n","Python3 program to calculate Pi using python long integers, BINARY\n","splitting and the Chudnovsky algorithm\n","\n","\"\"\"\n","\n","import math\n","from gmpy2 import mpz\n","from time import time\n","\n","def 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\n","check_digits = {\n","    100 : 70679,\n","    1000 :  1989,\n","    10000 : 75678,\n","    100000 : 24646,\n","    1000000 : 58151,\n","    10000000 : 55897,\n","}\n","\n","if __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","metadata":{},"outputs":[],"source":["import ast\n","import json\n","from urllib.request import urlopen, Request\n","\n","\n","# urllib请求解析,json\n","# url = 'https://ip8.com/ajax/resolve.php'\n","# request = Request(url)\n","# response = urlopen(request)\n","# content_dict = json.loads(response.read().decode('utf-8'))\n","\n","# eval\n","# url = 'https://ip8.com/ajax/resolve.php'\n","# request = Request(url)\n","# response = urlopen(request)\n","# content = eval(response.read().decode('utf-8'))\n","\n","# ast\n","url = 'https://ip8.com/ajax/resolve.php'\n","request = Request(url)\n","response = urlopen(request)\n","content_dict = ast.literal_eval(response.read().decode('utf-8'))\n","print(content_dict['resolved'])"]},{"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},"version":"3.7.3"},"orig_nbformat":2,"file_extension":".py","mimetype":"text/x-python","name":"python","npconvert_exporter":"python","pygments_lexer":"ipython3","version":3}}