矩阵的LU分解 c++

#include <iostream>
#include <math.h>
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;
}
原文地址:https://www.cnblogs.com/tiandsp/p/2271751.html