我如何排序方程式？

你需要乘法运算符：

X1 + X2 + 3 * X3

或者你想要那样：

X + X * X + 3 * X * X * X

您可以这样写：

X * ( 1 + X * ( 1 + 3 * X ) ) )

这里是C ++ 11中的另一个解决方案：

Here' one more solution in C++11:

#include <string>
#include <vector>
#include <cctype>
#include <iostream>
#include <algorithm>

typedef struct tagParam
{
	int coeff;
	int subscript;
} PARAM, *LPPARAM;

int main(int argc, char* argv[])
{
	// Let eq is the string containing the equation to be sorted
	std::string eq = "X2+X1+3X3+4X5+2X4+X7+X6";

	// Allocating a vector to store coeff + subscript objects
	std::vector<PARAM> params;
	// Parsing string containing the equation to be sorted
	for (auto FwdIt = eq.begin(); (FwdIt + 1) < eq.end(); FwdIt++)
	{
		// For each character perform a check if preceding character is coeff 
		// and succeeding character is subscript. 
		if (FwdIt != eq.end() && std::isdigit(*FwdIt) && \
			std::isdigit(*(FwdIt + 2)) && *(FwdIt + 1) == 'X')
		{
			// If so, append the coeff and subscript to lpParams buffer previously allocated
			PARAM param = { 0 };
			std::memset((void*)¶m, 0x00, sizeof(PARAM));
			param.coeff = *FwdIt - '0'; param.subscript = *(FwdIt + 2) - '0';
			if (FwdIt + 3 < eq.end()) FwdIt += 3; params.push_back(param);
		}

		// Perform a check if the current character is 'X' and next character is subscript
		else if (FwdIt != eq.end() && *FwdIt == 'X' && std::isdigit(*(FwdIt + 1)))
		{
			// If so, append subscript to lpParams buffer previously allocated
			PARAM param = { 0 };
			std::memset((void*)¶m, 0x00, sizeof(PARAM));
			param.subscript = *(FwdIt + 1) - '0'; params.push_back(param); 
			if (FwdIt + 2 < eq.end()) FwdIt += 2;

		}
	}

	// Perform a sort to order all coeff + subscript objects by the value of subscript
	std::sort(params.begin(), params.end(), [&](const PARAM& param1, \
		const PARAM& param2) { return param1.subscript < param2.subscript; });

	std::string output = "\0";
	// Constructing an output string
	for (auto It = params.begin(); It != params.end(); It++)
		// For each coeff + subscript object perform a check if the subscript
		// is not equal to 0 and coeff is equal to zero (there's no coeff)
		if (It->subscript != 0 && It->coeff == 0)
		{
			// If so, output X_subscript to string buffer
			output += "X" + std::to_string(It->subscript);
			if (It != params.end() - 1) output += "+";
		}
		
		// Otherwise, perform a check if both coeff and subscript are not zero
		else if (It->subscript != 0 && It->coeff != 0)
		{
			// if so, output coeff_X_subscript to string buffer
			output += std::to_string(It->coeff) + "X";
			output += std::to_string(It->subscript);
			if (It != params.end() - 1) output += "+";
		}

	// Output the resulting string buffer containing sorted equation
	std::cout << "input = " << eq << "\n" << "output = " << output;

	std::cin.get();

    return 0;
}

这是我的解决方案：

Here's my solution:

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <ctype.h>

typedef struct tagParam
{
	int coeff;
	int subscript;
} PARAM, *LPPARAM;

int main(int argc, char* argv[])
{
	// Let eq is a string buffer containing an equation to be parsed
	static char eq[256] = "X2+X1+3X3";

	// Allocate buffer for an array of subscript objects (coeff + subscript)
	LPPARAM lpParams = new PARAM[100];
	memset((void*)lpParams, 0x00, sizeof(PARAM) * 100);

	int n = 0;
	// Parsing string containing the equation to be sorted
	for (int i = 0; eq[i] != '\0'; i++)
	{
		// For each character perform a check if preceding character is coeff 
		// and succeeding character is subscript. 
		if (isdigit(eq[i]) && isdigit(eq[i + 2]) && eq[i + 1] == 'X')
		{
			// If so, append the coeff and subscript to lpParams buffer previously allocated
			lpParams[n].coeff = eq[i] - '0';
			lpParams[n++].subscript = eq[i + 2] - '0';
			i += 3;
		}

		// Perform a check if the current character is 'X' and next character is subscript
		else if (eq[i] == 'X' && isdigit(eq[i + 1]))
		{
			// If so, append subscript to lpParams buffer previously allocated
			lpParams[n++].subscript = eq[i + 1] - '0';
			i += 2;
		}
	}

	// Perform a sort to order all coeff + subscript objects by the value of subscript
	for (int i = 0; lpParams[i].subscript != 0; i++)
	{
		int min = i;
		for (int j = i + 1; lpParams[j].subscript != 0; j++)
			min = (lpParams[j].subscript < lpParams[min].subscript) ? j : min;

		PARAM temp = lpParams[i];
		lpParams[i] = lpParams[min];
		lpParams[min] = temp;
	}

	static char output[256] = "\0";
	// Constructing an output string
	for (int i = 0; lpParams[i].subscript != 0; i++)
		// For each object in the array perform a check if the subscript
		// is not equal to 0 and coeff is equal to zero (there's no coeff)
		if (lpParams[i].subscript != 0 && lpParams[i].coeff == 0)
			// If so, output X_subscript to string buffer
			sprintf_s(output, 256, "%sX%d+", output, lpParams[i].subscript);
		// Otherwise, perform a check if both coeff and subscript are not zero
		else if (lpParams[i].subscript != 0 && lpParams[i].coeff != 0)
			// if so, output coeff_X_subscript to string buffer
			sprintf_s(output, 256, "%s%dX%d+", output, lpParams[i].coeff, lpParams[i].subscript);

	// Shrink output buffer by one ending character (e.g. remove unnecessary '+' character)
	output[strlen(output) - 1] = '\0';

	// Output the resulting string buffer containing sorted equation
	printf("input = %s\noutput = %s\n", eq, output);

	_getch();

    return 0;
}