我如何只用基础知识来做到这一点 [英] How do I do this with only basics

问题描述

等式正确形式：任何操作都应该首先将每个等式转换为正确形式的

。等式本身应具有以下属性

1.变量按字母顺序从左到右排序

3x2 + 2x1 + 4x3 = 16

应该是

2x1 + 3x2 + 4x3 = 16

2.任何变量应该只出现一次

4x1 + 3x2-2x1 + 4x3 = 16

应该是

2x1 + 3x2 + 4x3 = 16

3.等式中只应出现一个常数项，它应该是

在右侧

2x1 + 3x2 + 5 + 4x3-11 = 10

应该是

2x1 + 3x2 + 4x3 = 16

第2页，共10页

4.等于1时的系数或-1数字1是可选的

1x1 + 3x2-1x3 = 10

可以输入为

x1 + 3x2-x3 = 10



< b>我尝试了什么：



我从第一点开始，将方程式分成另一个数组，但是我被卡住了我不知道该怎么办...任何帮助？

这是我尝试过的事情

PS我' m不允许使用复杂的方法或任何STL，只是数组，字符串，向量....任何基本的：/

Equation proper form : Any operation should first convert every equation to
the proper form. The equation proper should have the following properties
1. variables are ordered alphabetically from left to right
3x2+2x1+4x3=16
Should be
2x1+3x2+4x3=16
2. Any variable should appears only one time
4x1+3x2-2x1+4x3=16
Should be
2x1+3x2+4x3=16
3. Only one constant term should appear in the equation and it should be
on the right hand side
2x1+3x2+5+4x3-11=10
Should be
2x1+3x2+4x3=16
Page 2 of 10
4. Coefficient when equals to one or -1 the digit 1 is optional
1x1+3x2-1x3=10
Can be input as be
x1+3x2-x3=10

What I have tried:

I'm starting from the first point and splitting the equation and putting it into another array but I'm stuck and I don't know what should I do ... any help??
Here is what I have tried
PS I'm not allowed to use complex method or any STLs, just arrays, strings, vectors.... anything basic :/

cin >> s;
	for (int i = 0; i<s.size(); i++)
	{
		if (s[i] == '+')
		{
			copyy[a] += s[i];
			a++;
		}
		else
		{
			copyy[a] += s[i];
		}

	}

推荐答案

你的任务是等式，所以你必须找到所有x实例并将它们预先排序为一种子数组。这个子目标你必须评估，然后在最后重新构建解决方案。



这是令牌x1，x2等的一些字符串操作......



我想有些人不明白你的意思是x * x和x2; - ）

Your task is sort the equation, so you must find all x instances and pre-order them into an sub array of one kind. This subarry you must evaluate and than re-build the solution at end.

It is some string operations for the tokens "x1", "x2" and so on...

I guess some people dont understand that you mean "x*x" with "x2" ;-)

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