From cf239d8889b52e8ff4c855a3782998f056ee7d50 Mon Sep 17 00:00:00 2001 From: root Date: Tue, 15 Oct 2019 12:11:50 +0800 Subject: [PATCH] --- {LearnJava => JavaCode}/Jb01_HelloWorld.java | 0 {LearnJava => JavaCode}/Jb02_VarDemo1.java | 0 {LearnJava => JavaCode}/Jb03_VarDemo2.java | 0 {LearnJava => JavaCode}/Jb04_Operate2.java | 0 {LearnJava => JavaCode}/Jb05_Operate3.java | 0 {LearnJava => JavaCode}/Jb06_Operate4.java | 0 {LearnJava => JavaCode}/Jb07_IfDemo.java | 0 {LearnJava => JavaCode}/Jb08_IfTest.java | 0 {LearnJava => JavaCode}/Jb09_SwitchDemo.java | 0 {LearnJava => JavaCode}/Jb10_WhileDemo.java | 0 {LearnJava => JavaCode}/Jb11_WhileTest.java | 0 {LearnJava => JavaCode}/Jb12_ForDemo.java | 0 {LearnJava => JavaCode}/Jb13_ForTest.java | 0 {LearnJava => JavaCode}/Jb14_Function.java | 0 {LearnJava => JavaCode}/Jb15_Function2.java | 0 {LearnJava => JavaCode}/Jb16_Function3.java | 0 {LearnJava => JavaCode}/Jb17_ArrDemo.java | 0 {LearnJava => JavaCode}/Jb18_ArrSort.java | 0 {LearnJava => 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LearnPython3/AwesomeMath.py | 234 ------------------ Linux_man_cn/7z.md | 33 +-- Linux_man_cn/ls.md | 8 +- Py3Scripts/AwesomeMath.ipynb | 1 + {LearnPython3 => Py3Scripts}/GrabNovel.py | 0 {LearnPython3 => Py3Scripts}/ScanPort.py | 0 {LearnShell => ShellScripts}/LeapYear.sh | 0 {LearnShell => ShellScripts}/MaxDiskSpace.sh | 0 {LearnShell => ShellScripts}/ras_ro.sh | 0 50 files changed, 22 insertions(+), 254 deletions(-) rename {LearnJava => JavaCode}/Jb01_HelloWorld.java (100%) rename {LearnJava => JavaCode}/Jb02_VarDemo1.java (100%) rename {LearnJava => JavaCode}/Jb03_VarDemo2.java (100%) rename {LearnJava => JavaCode}/Jb04_Operate2.java (100%) rename {LearnJava => JavaCode}/Jb05_Operate3.java (100%) rename {LearnJava => JavaCode}/Jb06_Operate4.java (100%) rename {LearnJava => JavaCode}/Jb07_IfDemo.java (100%) rename {LearnJava => JavaCode}/Jb08_IfTest.java (100%) rename {LearnJava => JavaCode}/Jb09_SwitchDemo.java (100%) rename {LearnJava => JavaCode}/Jb10_WhileDemo.java (100%) rename 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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 - z = 100 - x - y # 小鸡花费z元为100元减去x和y - if x / 5 + y / 3 + z * 3 == 100: - print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3)) - # pass -# end = time.perf_counter_ns() -# time1 = end - start -# print("解法一花费时间：", time1) - -# 解法二：枚举法 -# 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 -# 之间不包括100 -for x in range(0, 20): - for y in range(0, 33): - z = 100 - x - y # 小鸡个数z等于100只减去公鸡x只加母鸡y只 - if 5 * x + 3 * y + z / 3 == 100: # 钱数相加等于100元 - print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x, y, z)) - -# 解法三：解法和解法一类似 -# 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 -# 少，100元不再是固定的。假设花费钱数依次为x、y、z元 -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 - 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 - for z in range(0, 101 - x - y): - if x / 5 + y / 3 + z * 3 == 100 and x + y + z == 100: # 花费和鸡数都是100 - print("公鸡：%d只，母鸡：%d只，小鸡：%d只" % (x / 5, y / 3, z * 3)) - -# 经典斐波那契数列 -# 定义:https://wikimedia.org/api/rest_v1/media/math/render/svg/c374ba08c140de90c6cbb4c9b9fcd26e3f99ef56 -# 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 -# 方法一：使用递归 -def fib1(n): - if n<0: - print("Incorrect input") - elif n==1: - return 0 # 第一个斐波那契数是0 - elif n==2: - return 1 # 第二斐波那契数是1 - else: - return fib1(n-1)+fib1(n-2) - -print(fib1(2)) - -# 方法二：使用动态编程 -FibArray = [0, 1] - - -def fib2(n): - if n < 0: - print("Incorrect input") - elif n <= len(FibArray): - return FibArray[n - 1] - else: - temp_fib = fib2(n - 1) + fib2(n - 2) - FibArray.append(temp_fib) - return temp_fib - -# 方法三：空间优化 - -def fibonacci(n): - a = 0 - b = 1 - if n < 0: - print("Incorrect input") - elif n == 0: - return a - elif n == 1: - return b - else: - for i in range(2,n): - c = a + b - a = b - b = c - return b - - -# 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 -# 找出1-1000之间的水仙花数 - -# 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 - -# 方法一：解答四个数组成不重复三位数(暂未想到更优方法) -for x in range(1, 5): - for y in range(1, 5): - for z in range(1, 5): - if (x != y) and (x != z) and (z != y): - print(x, y, z) - - -# 计算pi小数点任意位数 -from __future__ import division -import math -from time import time -time1 = time() -number = int(input('输入计算的位数：')) -number1 = number + 10 # 多计算十位方式尾数取舍影响 -b = 10 ** number1 - -# 求含4/5的首项 -x1 = b * 4 // 5 -# 求含1/239的首项 -x2 = b // -239 - -# 求第一大项 -he = x1 + x2 -# 设置下面循环的终点，即共计算n项 -number *= 2 - -# 循环初值=3，末值2n,步长=2 -for i in range(3, number, 2): - # 求每个含1/5的项及符号 - x1 //= -25 - # 求每个含1/239的项及符号 - x2 //= -57121 - # 求两项之和 - x = (x1 + x2) // i - # 求总和 - he += x - -# 求出π -pi = he * 4 -# 舍掉后十位 -pi //= 10 ** 10 - -# 输出圆周率π的值 -pi_string = str(pi) -result = pi_string[0] + str('.') + pi_string[1:len(pi_string)] -print(result) - -time2 = time() - -print(u'耗时：' + str(time2 - time1) + 's') - - -# 使用chudnovsky算法计算 -# 理解链接：https://www.craig-wood.com/nick/articles/pi-chudnovsky/ - -""" -Python3 program to calculate Pi using python long integers, BINARY -splitting and the Chudnovsky algorithm - -""" - -import math -from gmpy2 import mpz -from time import time - -def pi_chudnovsky_bs(digits): - """ - Compute int(pi * 10**digits) - - This is done using Chudnovsky's series with BINARY splitting - """ - C = 640320 - C3_OVER_24 = C**3 // 24 - def bs(a, b): - """ - Computes the terms for binary splitting the Chudnovsky infinite series - - a(a) = +/- (13591409 + 545140134*a) - p(a) = (6*a-5)*(2*a-1)*(6*a-1) - b(a) = 1 - q(a) = a*a*a*C3_OVER_24 - - returns P(a,b), Q(a,b) and T(a,b) - """ - if b - a == 1: - # Directly compute P(a,a+1), Q(a,a+1) and T(a,a+1) - if a == 0: - Pab = Qab = mpz(1) - else: - Pab = mpz((6*a-5)*(2*a-1)*(6*a-1)) - Qab = mpz(a*a*a*C3_OVER_24) - Tab = Pab * (13591409 + 545140134*a) # a(a) * p(a) - if a & 1: - Tab = -Tab - else: - # Recursively compute P(a,b), Q(a,b) and T(a,b) - # m is the midpoint of a and b - m = (a + b) // 2 - # Recursively calculate P(a,m), Q(a,m) and T(a,m) - Pam, Qam, Tam = bs(a, m) - # Recursively calculate P(m,b), Q(m,b) and T(m,b) - Pmb, Qmb, Tmb = bs(m, b) - # Now combine - Pab = Pam * Pmb - Qab = Qam * Qmb - Tab = Qmb * Tam + Pam * Tmb - return Pab, Qab, Tab - # how many terms to compute - DIGITS_PER_TERM = math.log10(C3_OVER_24/6/2/6) - N = int(digits/DIGITS_PER_TERM + 1) - # Calclate P(0,N) and Q(0,N) - P, Q, T = bs(0, N) - one_squared = mpz(10)**(2*digits) - sqrtC = (10005*one_squared).sqrt() - return (Q*426880*sqrtC) // T - -# The last 5 digits or pi for various numbers of digits -check_digits = { - 100 : 70679, - 1000 : 1989, - 10000 : 75678, - 100000 : 24646, - 1000000 : 58151, - 10000000 : 55897, -} - -if __name__ == "__main__": - digits = 100 - pi = pi_chudnovsky_bs(digits) - print(pi) - #raise SystemExit - for log10_digits in range(1,9): - digits = 10**log10_digits - start =time() - pi = pi_chudnovsky_bs(digits) - print("chudnovsky_gmpy_mpz_bs: digits",digits,"time",time()-start) - if digits in check_digits: - last_five_digits = pi % 100000 - if check_digits[digits] == last_five_digits: - print("Last 5 digits %05d OK" % last_five_digits) - else: - print("Last 5 digits %05d wrong should be %05d" % (last_five_digits, check_digits[digits])) \ No newline at end of file diff --git a/Linux_man_cn/7z.md b/Linux_man_cn/7z.md index 463aa3c..55c2c22 100644 --- a/Linux_man_cn/7z.md +++ b/Linux_man_cn/7z.md @@ -68,19 +68,21 @@ Usage: 7z [...] [...] p7zip p7zip-plugins ``` -# 支持的zip格式压缩方法 -* 0 - Store -* 1 - Shrink (decompression only) -* 6 - Implode (decompression only) -* 8 - Deflate -* 9 - Deflate64 -* 12 - BZip2 -* 14 - LZMA -* 95 - xz (decompression only) -* 98 - PPMd -* 99 - WinZip AES - -# 支持的7z格式压缩方法 +## 支持的zip格式压缩方法 + +* 0 - Store +* 1 - Shrink (decompression only) +* 6 - Implode (decompression only) +* 8 - Deflate +* 9 - Deflate64 +* 12 - BZip2 +* 14 - LZMA +* 95 - xz (decompression only) +* 98 - PPMd +* 99 - WinZip AES + +## 支持的7z格式压缩方法 + | 压缩方法 | 描述 | | :------: | :------: | | LZMA | Improved and optimized version of LZ77 algorithm | @@ -91,7 +93,7 @@ p7zip p7zip-plugins | BZip2 | Standard BWT algorithm | | Deflate | Standard LZ77-based algorithm | -# 退出码含义 +## 退出码含义 | Code | Meaning | | :------: | :------: | @@ -123,7 +125,6 @@ p7zip p7zip-plugins ## 恢复损坏的7z压缩包 错误类型有两种： + * 解压或测试时，提示：数据错误或CRC错误 * 打开时，提示：文件'a.7z'作为存档无法打开 - - diff --git a/Linux_man_cn/ls.md b/Linux_man_cn/ls.md index 6261ed0..0b8361e 100644 --- a/Linux_man_cn/ls.md +++ b/Linux_man_cn/ls.md @@ -10,18 +10,18 @@ -C 多列显示输出结果。这是默认选项 -l 与“-C”选项功能相反，所有输出信息用单列格式输出，不输出为多列 -F 在每个输出项后追加文件的类型标识符，具体含义 * 表示具有可执行权限的普通文件，“/”表示目录，“@”表示符 - 号链接，“|”表示命令管道FIFO，“=”表示sockets套接字。当文件为普通文件时，不输出任何标识符 + 号链接，“|”表示命令管道FIFO，“=”表示sockets套接字。当文件为普通文件时，不输出任何标识符 -b 将文件中的不可输出的字符以反斜线“”加字符编码的方式输出 -c 与“-lt”选项连用时，按照文件状态时间排序输出目录内容，排序的依据是文件的索引节点中的ctime字段。与“-l” - 选项连用时，则排序的一句是文件的状态改变时间 + 选项连用时，则排序的一句是文件的状态改变时间 -d 仅显示目录名，而不显示目录下的内容列表。显示符号链接文件本身，而不显示其所指向的目录列表 -f 此参数的效果和同时指定“aU”参数相同，并关闭“lst”参数的效果 -i 显示文件索引节点号（inode）。一个索引节点代表一个文件 --file-type 与“-F”选项的功能相同，但是不显示“*” -k 以KB（千字节）为单位显示文件大小 -h 将文件容量以人类易读的方式显示 --l 以长格式显示目录下的内容列表。输出的信息从左到右依次包括文件名，文件类型、权限模式、硬连接数、所有者、组 - 、文件大小和文件的最后修改时间等 +-l 以长格式显示目录下的内容列表。输出的信息从左到右依次包括文件名，文件类型、权限模式、硬连接数、所有者、组、文件大小 + 和文件的最后修改时间等 -m 用“,”号区隔每个文件和目录的名称 -n 以用户识别码和群组识别码替代其名称 -r 以文件名反序排列并输出目录内容列表 diff --git a/Py3Scripts/AwesomeMath.ipynb b/Py3Scripts/AwesomeMath.ipynb new file mode 100644 index 0000000..f1508a1 --- /dev/null +++ b/Py3Scripts/AwesomeMath.ipynb @@ -0,0 +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}} \ No newline at end of file diff --git a/LearnPython3/GrabNovel.py b/Py3Scripts/GrabNovel.py similarity index 100% rename from LearnPython3/GrabNovel.py rename to Py3Scripts/GrabNovel.py diff --git a/LearnPython3/ScanPort.py b/Py3Scripts/ScanPort.py similarity index 100% rename from LearnPython3/ScanPort.py rename to Py3Scripts/ScanPort.py diff --git a/LearnShell/LeapYear.sh b/ShellScripts/LeapYear.sh similarity index 100% rename from LearnShell/LeapYear.sh rename to ShellScripts/LeapYear.sh diff --git a/LearnShell/MaxDiskSpace.sh b/ShellScripts/MaxDiskSpace.sh similarity index 100% rename from LearnShell/MaxDiskSpace.sh rename to ShellScripts/MaxDiskSpace.sh diff --git a/LearnShell/ras_ro.sh b/ShellScripts/ras_ro.sh similarity index 100% rename from LearnShell/ras_ro.sh rename to ShellScripts/ras_ro.sh