Learn/Py3Scripts/AwesomePython.ipynb

{"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