58 lines
1.4 KiB
C++
58 lines
1.4 KiB
C++
#include <iostream>
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#include <vector>
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#include <cmath>
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extern "C" {
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void dgetrf_(int* m, int* n, double* a, int* lda, int* ipiv, int* info);
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void dgetri_(int* n, double* a, int* lda, int* ipiv, double* work, int* lwork, int* info);
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}
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int main() {
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const int N = 3;
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double A_data[] = {
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1, 0, 5, // 第1列
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2, 1, 6, // 第2列
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3, 4, 0 // 第3列
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};
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std::vector<double> A(A_data, A_data + 9);
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std::cout << "Original matrix (column-major):\n";
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for (int i = 0; i < N; ++i) {
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for (int j = 0; j < N; ++j) {
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std::cout << A[j*N + i] << " ";
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}
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std::cout << "\n";
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}
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// LU 分解
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std::vector<int> ipiv(N);
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int info;
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int n = N;
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int lda = N;
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dgetrf_(&n, &n, &A[0], &lda, &ipiv[0], &info);
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if (info != 0) {
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std::cerr << "LU decomposition failed! (info=" << info << ")\n";
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return 1;
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}
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// 计算逆矩阵
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std::vector<double> work(N);
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int lwork = N; // 关键：必须是可变变量
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dgetri_(&n, &A[0], &lda, &ipiv[0], &work[0], &lwork, &info);
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if (info != 0) {
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std::cerr << "Matrix inversion failed! (info=" << info << ")\n";
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return 1;
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}
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std::cout << "\nInverse matrix:\n";
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for (int i = 0; i < N; ++i) {
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for (int j = 0; j < N; ++j) {
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std::cout << A[j*N + i] << " ";
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}
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std::cout << "\n";
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}
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return 0;
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} |