我如何排序方程式？

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具有X2 + X1 + 3X3 ----> X1 + X2 + 3X3 我的尝试： 我尝试分裂，但显然我做错了所以任何帮助？

having X2+X1+3X3 ----> X1+X2+3X3 What I have tried: I tried splitting but apparently I'm doing something wrong so any help?

推荐答案

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; }