说明：   void qr(double *a, double *d, int n) 矩阵的QR分解 假设数组an*n在内存中按行优先次序存放，此函数使用HouseHolder变换将其就地进行QR分解。 d为输出参数，d[0,n)存放QR分解的上三角矩阵对角线元素。 bool householder(double const *qr, double const *d, double *b, int n) 求线性代数方程组的解。 假设矩阵qrn*n为某个矩阵an*n的QR分解，在内存中按行优先次序存放。d[0,n)为其QR分解上三角矩阵对角线元素，b为方程组Ax=b的右端向量。 函数计算方程组Ax=b的解，并将结果存放在数组b[0,n)中。 函数成功时返回false，否则返回true.
(void qr (double* a, double* d, int n) matrix QR decomposition assuming an* n array in memory stored in row order, this function uses HouseHolder transform the place for QR decomposition. d as the output parameter, d [0, n) storage QR decomposition of the diagonal elements of upper triangular matrices. bool householder (double const* qr, double const* d, double* b, int n) of linear algebraic equations. Suppose matrix qrn* n for a matrix of an* n of the QR decomposition, in memory stored in row order. d [0, n) for QR decomposition of the triangular matrix diagonal elements, b for the equations Ax = b of the right end of the vector. Function calculates the equations Ax = b of the solution, and the results stored in array b [0, n) in. Function successfully returns false, otherwise true.)