Learn/Py3Scripts/AwesomeMath.ipynb

{"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}}