计算N阶行列式

根据N阶行列式计算公式：

$\sum _{j1,j2...,jn}(-1{)}^{\tau (j_1{j}_2...j_n)}{a}_{1j_1}{a}_{2j_2}...a_{n{j}_n}$

将j1,j2,…,jn进行全排列并且计算出他们的反序数，通过循环将各项累加即可

#include 
#include 
#include 
/*计算行列式
data：行列式的所有数据
n:行列式的阶数*/
double CalcDet(double *data, int n)
{
	std::vector<int> order;//序列的顺序
	double sum = 0;//累加的结果
	auto GetNum=[](std::vector<int> &a)
	{
		int n = a.size() - 1;//最后一个数不用管
		int sum = 0;
		for (int i = 0; i < n; ++i)
			for (int j = i + 1; j <= n; ++j)
					++sum;
		return sum;
	};//获取反序列数目的lambda表达式

	//初始化序列
	for (int i = 0; i < n; ++i)
		order.push_back(i);

	do{
		double mul = 1.0;
		for (int i = 0; i < n; ++i)
			mul *= data[i + order[i] * n];
		sum += pow(-1, GetNum(order))*mul;
	} while (std::next_permutation(order.begin(), order.end()));

	return sum;
}

测试代码：

#include 
using namespace std;
int main()
{
	double *data;
	int n;

	cout << "请输入阶数：";
	cin >> n;

	data = new double[n*n];

	cout << "请输入数据（空格或者换行隔开）" << endl;
	for (int i = 0; i < n*n; ++i)
		cin >> data[i];

	cout << "计算结果为：" << CalcDet(data, n) << endl;
	return 0;
}