Gauss-Jordan消元法求矩阵的逆

　　最近信息安全作业要写hill密码，涉及到求矩阵的逆。不想用一个函数就用那些乱七八糟的库，大概找了找又没现成的，只能自己撸了。

　　写的时候才发现线代都忘光了QAQ

　　

 1 //Gauss-Jordan Elimination method to get inverse matrix
 2 template double>
 3 std::vectordouble>> matrixInversion(const std::vector>& a)
 4 {
 5     const static double eps = 1e-6;
 6     std::vectordouble>> augma; //augmented matrix
 7     for (auto& vec : a)
 8     {
 9         std::vector<double> tmp;
10         std::transform(vec.begin(), vec.end(), std::back_inserter(tmp), [](const T interger)
11             {
12                 return static_cast<double>(interger);
13             });
14         augma.push_back(std::move(tmp));
15     }
16     int rank = a.size();
17     //init 
18     for (int i = 0; i < rank; ++i)
19     {
20         augma[i].resize(rank * 2);
21         augma[i][rank + i] = 1;
22     }
23     for (int i = 0; i < rank; ++i)
24     {
25         if (::abs(augma[i][i]) < eps)
26         {
27             int j;
28             for (j = i + 1; j < rank; ++j)
29             {
30                 if (::abs(augma[j][i]) > eps) break;
31             }
32             if (j == rank) return {};
33             for (int r = i; r < 2 * rank; ++r)
34             {
35                 augma[i][r] += augma[j][r];
36             }
37         }
38         double ep = augma[i][i];
39         for (int r = i; r < 2 * rank; ++r)
40         {
41             augma[i][r] /= ep;
42         }
43 
44         for (int j = i + 1; j < rank; ++j)
45         {
46             double e = -1 * (augma[j][i] / augma[i][i]);
47             for (int r = i; r < 2 * rank; ++r)
48             {
49                 augma[j][r] += e * augma[i][r];
50             }
51         }
52     }
53     for (int i = rank - 1; i >= 0; --i)
54     {
55         for (int j = i - 1; j >= 0; --j)
56         {
57             double e = -1 * (augma[j][i] / augma[i][i]);
58             for (int r = i; r < 2 * rank; ++r)
59             {
60                 augma[j][r] += e * augma[i][r];
61             }
62         }
63     }
64 
65     for (auto& vec : augma)
66     {
67         vec.erase(vec.begin(), vec.begin() + rank);
68     }
69 
70     return augma;
71 }