化学方程式自动配平的C++
时间: 2023-07-11 15:39:59 浏览: 60
好的,这里是一个简单的 C++ 代码示例,可以用于自动配平化学方程式:
```cpp
#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <algorithm>
using namespace std;
struct Element {
string name;
int count;
};
bool operator<(const Element& a, const Element& b) {
return a.name < b.name;
}
class Equation {
private:
vector<Element> reactants, products;
map<string, int> elementCounts;
void parseFormula(vector<Element>& elements, const string& formula, int& pos) {
while (pos < formula.size()) {
int i = pos + 1;
while (i < formula.size() && islower(formula[i])) ++i;
string name = formula.substr(pos, i - pos);
int count = 0;
while (i < formula.size() && isdigit(formula[i])) {
count = count * 10 + (formula[i] - '0');
++i;
}
if (count == 0) count = 1;
elements.push_back({name, count});
pos = i;
}
}
void countElements() {
for (auto& element : reactants) {
elementCounts[element.name] -= element.count;
}
for (auto& element : products) {
elementCounts[element.name] += element.count;
}
}
public:
void addReactant(const string& formula) {
int pos = 0;
parseFormula(reactants, formula, pos);
}
void addProduct(const string& formula) {
int pos = 0;
parseFormula(products, formula, pos);
}
void balance() {
countElements();
vector<Element> elements;
for (auto& [name, count] : elementCounts) {
elements.push_back({name, count});
}
sort(elements.begin(), elements.end());
int n = elements.size();
vector<vector<int>> matrix(n, vector<int>(n + 1));
for (int i = 0; i < n; ++i) {
matrix[i][n] = -elements[i].count;
for (int j = 0; j < n; ++j) {
for (auto& element : (i < j ? reactants : products)) {
if (element.name == elements[j].name) {
matrix[i][j] += element.count;
}
}
}
}
for (int i = 0; i < n; ++i) {
int k = i;
for (int j = i + 1; j < n; ++j) {
if (abs(matrix[j][i]) > abs(matrix[k][i])) {
k = j;
}
}
if (k != i) {
swap(matrix[i], matrix[k]);
}
for (int j = i + 1; j <= n; ++j) {
matrix[i][j] /= matrix[i][i];
}
for (int j = 0; j < n; ++j) {
if (j != i) {
for (int k = i + 1; k <= n; ++k) {
matrix[j][k] -= matrix[j][i] * matrix[i][k];
}
}
}
}
for (int i = 0; i < n; ++i) {
int factor = matrix[i][n];
for (auto& element : (i < n / 2 ? reactants : products)) {
if (element.name == elements[i].name) {
element.count *= factor;
}
}
}
}
void print() const {
for (auto& element : reactants) {
cout << element.count << element.name << " + ";
}
cout << "\b\b -> ";
for (auto& element : products) {
cout << element.count << element.name << " + ";
}
cout << "\b\b" << endl;
}
};
int main() {
Equation equation;
equation.addReactant("H2");
equation.addReactant("O2");
equation.addProduct("H2O");
equation.balance();
equation.print();
return 0;
}
```
这个示例代码实现了化学方程式的自动配平功能,可以根据给定的反应物和生成物,计算出平衡后的反应方程式。具体实现过程是:
1. 将反应物和生成物分别解析为元素及其个数的列表;
2. 根据元素个数的守恒原则,计算出元素的总个数,并将其存储到一个映射表中;
3. 构造一个线性方程组,将元素的个数表示为未知数,并将方程组系数矩阵填充;
4. 使用高斯-约旦消元法求解方程组,得到每个元素的系数;
5. 根据每个元素的系数,计算出平衡后的反应方程式。
当然,这只是一个简单的示例代码,实际的化学方程式自动配平程序还需要考虑更多的细节和特殊情况。