C++实现lu分解

【文件属性】：
文件名称：C++实现lu分解
文件大小：1KB
文件格式：TXT
更新时间：2018-12-18 14:57:10
C++，lu #include #include using namespace std; int main() { double a[] = { 4, 2, -2, 2, 2, -3, -2, -3, 14 }; double *L = NULL; double *U = NULL; double tmp=0; int n; //矩阵内总数据个数 int s; //矩阵的阶数 n = sizeof(a)/sizeof(double); s = sqrtl(n); L = new double[n]; U = new double[n]; for (int i = 0; i < s; i++) { for (int j = 0; j < s; j++) { if (i == j) L[i*s+j] = 1; if (i < j) L[i*s+j] = 0; if (i > j) U[i*s+j] = 0; U[0*s+j] = a[0*s+j]; L[i*s+0] = a[i*s+0] / U[0*s+0]; } } for (int k = 1; k < s; k++) { for (int j = k; j < s; j++) { tmp = 0; for (int m = 0; m < k; m++) { tmp += L[k*s+m] * U[m*s+j]; } U[k*s+j] = a[k*s+j] - tmp; } for (int i = k+1; i < s; i++) { tmp = 0; for (int m = 0; m < k; m++) { tmp += L[i*s+m] * U[m*s+k]; } L[i*s+k] = ( a[i*s+k] - tmp ) / U[k*s+k]; } } //这里就得到L和U矩阵的值了 delete [] L; delete [] U; system("pause"); return 0; }