python的除法进一位怎样

Python 实现带有计算过程的拓展欧几里得除法

何为拓展欧几里得除法，网上的资料一搜便有。并且，拓展欧几里得除法的代码也是随处可见。然而，笔者比较贪心，不仅想要结果，还想要把计算过程让机器一并处理出来以方便抄写。甚至，有些本科课程要求使用多项式进行运算，所以贴出此代码以供大家参考。

由于本人不是计算机专业，也不是数学专业，且该代码经历了“从无过程到有欧几里得除法过程，再到有拓展欧几里得除法过程”的三次革命，本代码有诸多不足，例如这种写法可能不是直接给出过程的最好写法，过程输出没有宏观进行调整。另外，命令行模式中的拓展欧几里得除法部分采用的不是本科课本上常见的计算步骤，但它也能得出正确答案。

from os import system
from sys import argv, exit, stdin
from msvcrt import getch
PLATFORM = __import__("platform").system().lower()
EXIT_SUCCESS = 0
EXIT_FAILURE = 1
pars = []


def clearScreen(fakeClear = 120) -> None:
	if stdin.isatty():#在终端
		if "windows" == PLATFORM:
			system("cls")
		elif "linux" == PLATFORM:
			system("clear")
		else:
			try:
				print("\n" * int(fakeClear))
			except:
				print("\n" * 120)
	else:
		try:
			print("\n" * int(fakeClear))
		except:
			print("\n" * 120)

def getMulMode(x, mul_mode = None) -> str:
	if mul_mode is None:
		return str(x)
	elif type(mul_mode) == int and mul_mode > 0:
		return str(x) + " " * (mul_mode - 1) + "×"
	elif type(mul_mode) == int and mul_mode < 0:
		return "×" + " " * (-mul_mode - 1) + str(x)
	else:
		return ""

def getX(n, mul_mode = None) -> str: # None = 默认，0 = 隐藏，1 = 右输出" × "，-1 = 左输出" × "
	if type(n) in (tuple, list, set):
		return [getX(item, mul_mode) for item in n]
	elif type(n) == dict:
		return {item:getX(item, mul_mode) for item in list(n.keys())}
	if type(n) != int:
		return str(n)
	elif n < 0:
		return "-" + getX(-n)
	in_list = bin(n)[2:]
	out_list = []
	for i, ele in enumerate(in_list[:-2]):
		if ele == "1":
			out_list.append("x^{0}".format(len(in_list) - i - 1))
	else:
		if len(in_list) > 1 and in_list[-2] == "1":
			out_list.append("x")
		if in_list[-1] == "1":
			out_list.append("1")
	if out_list:
		if len(out_list) == 1:
			if out_list == ["1"]:
				return getMulMode("1", mul_mode)
			else:
				return out_list[0]
		else:
			return "(" + " + ".join(out_list) + ")"
	else:
		return getMulMode("0", mul_mode)

def getDHX(n) -> str: # 连等号输出
	if type(n) in (tuple, list, set):
		return [getDHX(item) for item in n]
	elif type(n) == dict:
		return {item:getDHX(item) for item in list(n.keys())}
	elif type(n) == int:
		return " = ".join([str(n), hex(n), getX(n)])
	else:
		return str(n)

def gcd(m, n, ll = None, isPrintX = False) -> int:# 定义 gcd 函数
	if ll is None:
		l = max(len(getX(m)), len(getX(n))) if isPrintX else len(str(max(abs(m), abs(n))))
		global pars
		pars = []
	else:
		l = ll
	try:
		if m == 0 or n == 0:
			return 0
		else:
			p = max(abs(m), abs(n))
			q = min(abs(m), abs(n))
			if l != 0: # 为输出而设立
				print("\t", " " * (l - len(str(p))), p, "=", q, "×", p // q, "＋", p % q)
				if isPrintX:
					print("\t", " " * (l - len(getX(p))), getX(p), "=", (getX(q, 1) + getX(p // q, -1)).replace("×  ×", "×") , "＋", getX(p % q))
				pars.append([p, q, p // q, p % q])
			if p % q == 0:
				return q
			else:	
				return gcd(q, p % q, l)
	except:
		return -2

def exgcd(p, q) -> tuple:# 定义exgcd 函数
	if p == 0 and q != 0:
		return "任意整数", 1
	elif q == 0 and p != 0:
		return 1, "任意整数"
	elif p == 0 and q == 0:
		return "任意整数", "任意整数"
	m = max(abs(p), abs(q))
	n = min(abs(p), abs(q))
	x = j = 1
	y = i = 0
	c = m
	d = n
	o = c // d
	r = c % d
	while r:
		c = d
		d = r
		z = x
		x = i
		i = z - o * i
		z = y
		y = j
		j = z - o * j
		o = c // d
		r = c % d
	if abs(p) >= abs(q):
		return (i if p >= 0 else -i), (j if q >= 0 else -j)
	else:
		return (j if p >= 0 else -j), (i if q >= 0 else -i)

def printProcess(isPrintX = False) -> None:
	if len(pars) == 1:
		print("\t", pars[0][1], "=", pars[0][1], "×", 1, "－", pars[0][0], "×", 0)
		return
	elif len(pars) < 1:
		return
	length = len(getX(pars[-1][1])) if isPrintX else len(str(pars[-1][1]))
	a = pars[-2][0]; aa = None
	b = 1
	c = pars[-2][1]; cc = None
	d = pars[-2][2]
	print("\t", " " * (length - len(str(pars[-1][1]))), pars[-1][1], "=", a, "×", b, "－", c, "×", d)
	if isPrintX:
		print("\t", " " * (length - len(getX(pars[-1][1]))), getX(pars[-1][1]), "=", (getX(a, 1) + getX(b, -1)).replace("×  ×", "×"), "－", (getX(c, 1) + getX(d, -1)).replace("×  ×", "×"))
	length += 1#修正上方因多一个参数导致多余一个空格
	for line in range(len(pars) - 3, -1, -1):
		if cc is None:
			aa = None # Swap
			cc = [pars[line][0], 1, pars[line][1], pars[line][2]]
			print("\t", " " * length, "=", a, "×", b, "－", "({0} × {1} － {2} × {3})".format(*cc), "×", d)
			if isPrintX:
				print("\t", " " * length, "=", (getX(a, 1) + getX(b, -1)).replace("×  ×", "×"), "－", "({0}{1} － {2}{3})".format(getX(cc[0], 1), getX(cc[1], -1), getX(cc[2], 1), getX(cc[3], -1)).replace("×  ×", "×") + getX(d, -1))
			c = cc[0]
			b += cc[3] * d
			d *= cc[1]
		else:
			cc = None # Swap
			aa = [pars[line][0], 1, pars[line][1], pars[line][2]]
			print("\t", " " * length, "=", "({0} × {1} － {2} × {3})".format(*aa), "×", b, "－", c, "×", d)
			if isPrintX:
				print("\t", " " * length, "=", "({0}{1} － {2}{3})".format(getX(aa[0], 1), getX(aa[1], -1), getX(aa[2], 1), getX(aa[3], -1)).replace("×  ×", "×") + getX(b, -1), "－", (getX(c, 1) + getX(d, -1)).replace("×  ×", "×"))
			a = aa[0]
			d += aa[3] * b
			b *= aa[1]
		print("\t", " " * length, "=", a, "×", b, "－", c, "×", d)
		if isPrintX:
			print("\t", " " * length, "=", (getX(a, 1) + getX(b, -1)).replace("×  ×", "×"), "－", (getX(c, 1) + getX(d, -1)).replace("×  ×", "×"))			
	print()
	
def main() -> int:
	if 3 == len(argv) or 4 == len(argv):
		try:
			a = int(argv[1])
			b = int(argv[2])
			if 4 == len(argv) and ("Y" == argv[3].upper() or "/Y" == argv[3].upper()):
				print("gcd({0}，{1}) = {2}".format(getDHX(a), getDHX(b), getDHX(gcd(a, b, 0))))
				print("({0}，{1})".format(*getDHX(exgcd(a, b))))
			else:
				print("gcd({0}，{1}) = {2}".format(a, b, gcd(a, b, 0)))
				print("({0}，{1})".format(*exgcd(a, b)))
			return EXIT_SUCCESS
		except:
			print("命令行参数不正确，请确保两个给定的数是（正）整数！")
			return EXIT_FAILURE
	else:
		if stdin.isatty():
			system("title 拓展欧几里得除法&color e")
		clearScreen()
		while True:
			while True:
				clearScreen()
				try:
					a = int(input("请输入整数 a："))
					break
				except:
					print("\a错误：输入的数不是整数，请按任意键重新输入。")
					getch()
			while True:
				clearScreen()
				try:
					print("a =", a)
					b = int(input("请输入整数 b："))
					break
				except:
					print("\a错误：输入的数不是整数，请按任意键重新输入。")
					getch()
			isPrintX = (input("是否启用多项式打印[y/N]？").upper() == "Y")
			clearScreen()
			print("正在使用欧几里得除法运行计算 gcd({0}, {1})，运算过程如下：".format(a, b))
			result = gcd(a, b, isPrintX = isPrintX)
			if result == -2:
				clearScreen()
				print("gcd(a = {0}，b = {1})".format(a, b))
				print("\a运算出错，可能存在算术溢出。")
			else:
				if result == 0:
					clearScreen()
					if isPrintX:
						print("a = {0} 和 b = {1} 的最大公因数为 {2}。".format(getDHX(a), getDHX(b), getDHX(max(abs(a), abs(b)))))
					else:
						print("a = {0} 和 b = {1} 的最大公因数为 {2}。".format(a, b, max(abs(a), abs(b))))
				else:
					if isPrintX:
						print("\n经运算，a = {0} 和 b = {1} 的最大公因数为 {2}。".format(getDHX(a), getDHX(b), getDHX(result)))
					else:
						print("\n经运算，a = {0} 和 b = {1} 的最大公因数为 {2}。".format(a, b, result))
				s, t = exgcd(a, b)
				if type(s) == int and type(t) == int:
					print("正在使用拓展欧几里得除法求 s、t，运算过程如下：")
					printProcess(isPrintX = isPrintX)
				if isPrintX:
					print("\n拓展欧几里得除法得 s = {0}，t = {1}。".format(getDHX(s), getDHX(t)))
				else:
					print("\n拓展欧几里得除法得 s = {0}，t = {1}。".format(s, t))
			print("\n\n\n输入“exit”（不区分大小写）回车退出程序，输入其它或直接回车运行新运算。")
			if "exit" == input("").lower():
				clearScreen()
				break
		return EXIT_SUCCESS



if __name__ == "__main__":
	exit(main())