python 求解一元二次方程组

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我是靠谱客的博主满意眼神，这篇文章主要介绍python 求解一元二次方程组，现在分享给大家，希望可以做个参考。

最近在做笔试题的时候遇到的一道题，挺有意思的贴出来和大佬讨论一下（常规的做法，不保证全对）

题目要求：

给两个方程，字符串格式给出，保证系数为整数。如6x-7+3y=2-5x和
-5+x+y=8-x，保证有解且为整数。
输入：两个方程；输出x,y的解，列表形式给出。如[1,2]表示x=1，y=2。

整体思路：
字符串提取x,y,常数项( c )的系数，使用哈希表实现
考虑到给定的方程需要化简，因此第一步要化简等号左右两端的方程，并获取系数，并存放在字典中，移到等式一端，之后合并字典则6x-7+3y=2-5x化简结果为{x:11, y:3, c:-9}
拿得到的系数使用消元法求解最终解

代码如下：该实现共包含三个函数：
transDict：用于获取参数，把给定的6x-7+3y=2-5x等号两端化简成两个字典
simplify：化简方程，调用transDict，化简整个参数为一个包含了整个方程系数的字典。
linearEquation：求解方程，消元法

复制代码

import collections

class Solution:
    def transDict(self, equation):
        dictE = collections.defaultdict(int) # 默认字典类型，key：x,y,c对应的value为整形
        sign, value = 1, 0 #一个符号标志，一个值标志

if equation[0] == '-': # 判断第一个是否是负号
            start = 1
            sign = -1
        else:
            start = 0

size = len(equation)
        for i in range(start, size):
            if equation[i] in '-+': # 第二个负号，参数写入字典
                value *= sign
                sign = 1
                dictE['c'] += value # 常数项系数
                value = 0
                if equation[i] == '-':# 提取下一个符号
                    sign = -1

elif equation[i] in 'xy': # 提取x,y系数
                if not value: # x,y前面没系数，设为1
                    value = 1
                value *= sign
                sign = 1

if equation[i] == 'x': # 更新字典
                    dictE['x'] += value
                if equation[i] == 'y':
                    dictE['y'] += value
                value = 0

elif equation[i].isdigit(): # 大于9的数加和
                value = value * 10 + int(equation[i])

if i == size - 1:
                    value *= sign
                    dictE['c'] += value
        return dictE

def simplify(self, equation):
        equation = equation.split('=')
        leftE, rightE = equation[0], equation[1]

dictLeft = self.transDict(leftE) # 得到等式左边的字典
        dictRight = self.transDict(rightE) # 得到等式右边的字典

for key, val in dictLeft.items(): # 合并
            dictLeft[key] -= dictRight[key]

return dictLeft

def linearEquation(self, equation1, equation2): # 消元法求解
        eDict1 = self.simplify(equation1)
        eDict2 = self.simplify(equation2)
        if eDict1['y'] != eDict2['y']:
            value = eDict2['y'] / eDict1['y']
            eDict1['y'] = eDict2['y']
            eDict1['x'] *= value
            eDict1['x'] -= eDict2['x']
            eDict1['c'] *= value
            eDict1['c'] -= eDict2['c']

x = -eDict1['c'] / eDict1['x']
        y = (-eDict2['c'] - eDict2['x'] * x) // eDict2['y']
        return [int(x), int(y)]
    
''' 
这里也可以用[线性代数](https://baike.baidu.com/item/%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%BB%84/7594389#2)知识求解
ax+by=e
cx+dy=f

x = (d*e-b*f)/(ad-bc)
y = (a*f-c*e)/(ad-bc)
'''

if __name__ == '__main__':
    s = Solution()
    while 1: # 循环测试
        equation1 = input()
        equation2 = input()
        print(s.linearEquation(equation1, equation2))
'''
x+3y=9
2x+y=13
6x-7+3y=2-5x
-5+x+y=8-x
'''

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import collections

class Solution:
    def transDict(self, equation):
        dictE = collections.defaultdict(int) # 默认字典类型，key：x,y,c对应的value为整形
        sign, value = 1, 0 #一个符号标志，一个值标志

        if equation[0] == '-': # 判断第一个是否是负号
            start = 1
            sign = -1
        else:
            start = 0

        size = len(equation)
        for i in range(start, size):
            if equation[i] in '-+': # 第二个负号，参数写入字典
                value *= sign
                sign = 1
                dictE['c'] += value # 常数项系数
                value = 0
                if equation[i] == '-':# 提取下一个符号
                    sign = -1

            elif equation[i] in 'xy': # 提取x,y系数
                if not value: # x,y前面没系数，设为1
                    value = 1
                value *= sign
                sign = 1

                if equation[i] == 'x': # 更新字典
                    dictE['x'] += value
                if equation[i] == 'y':
                    dictE['y'] += value
                value = 0

            elif equation[i].isdigit(): # 大于9的数加和
                value = value * 10 + int(equation[i])

                if i == size - 1:
                    value *= sign
                    dictE['c'] += value
        return dictE

    def simplify(self, equation):
        equation = equation.split('=')
        leftE, rightE = equation[0], equation[1]

        dictLeft = self.transDict(leftE) # 得到等式左边的字典
        dictRight = self.transDict(rightE) # 得到等式右边的字典

        for key, val in dictLeft.items(): # 合并
            dictLeft[key] -= dictRight[key]

        return dictLeft

    def linearEquation(self, equation1, equation2): # 消元法求解
        eDict1 = self.simplify(equation1)
        eDict2 = self.simplify(equation2)
        if eDict1['y'] != eDict2['y']:
            value = eDict2['y'] / eDict1['y']
            eDict1['y'] = eDict2['y']
            eDict1['x'] *= value
            eDict1['x'] -= eDict2['x']
            eDict1['c'] *= value
            eDict1['c'] -= eDict2['c']

        x = -eDict1['c'] / eDict1['x']
        y = (-eDict2['c'] - eDict2['x'] * x) // eDict2['y']
        return [int(x), int(y)]
    
''' 
这里也可以用[线性代数](https://baike.baidu.com/item/%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%BB%84/7594389#2)知识求解
ax+by=e
cx+dy=f

x = (d*e-b*f)/(ad-bc)
y = (a*f-c*e)/(ad-bc)
'''

if __name__ == '__main__':
    s = Solution()
    while 1: # 循环测试
        equation1 = input()
        equation2 = input()
        print(s.linearEquation(equation1, equation2))
'''
x+3y=9
2x+y=13
6x-7+3y=2-5x
-5+x+y=8-x
'''