ldl分解求解方程组 c语言
时间: 2023-12-21 15:30:17 浏览: 29
以下是使用LDL分解法求解线性方程组的C语言代码示例:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define N 3
void ldl(double A[][N], double L[][N], double D[], int n) {
int i, j, k;
double sum;
for (i = 0; i < n; i++) {
for (j = 0; j <= i; j++) {
sum = A[i][j];
for (k = 0; k < j; k++) {
sum -= L[i][k] * D[k] * L[j][k];
}
if (i == j) {
D[i] = sum;
L[i][i] = 1;
} else {
L[i][j] = sum / D[j];
}
}
for (j = i + 1; j < n; j++) {
sum = A[j][i];
for (k = 0; k < i; k++) {
sum -= L[j][k] * D[k] * L[i][k];
}
L[j][i] = sum / D[i];
}
}
}
void solve(double L[][N], double D[], double b[], double x[], int n) {
int i, j;
double y[N], sum;
for (i = 0; i < n; i++) {
sum = b[i];
for (j = 0; j < i; j++) {
sum -= L[i][j] * y[j];
}
y[i] = sum / L[i][i];
}
for (i = 0; i < n; i++) {
x[i] = y[i] / sqrt(D[i]);
}
for (i = n - 1; i >= 0; i--) {
sum = x[i];
for (j = i + 1; j < n; j++) {
sum -= L[j][i] * x[j];
}
x[i] = sum / L[i][i];
}
}
int main() {
double A[N][N] = {{4, -2, 2}, {-2, 2, -4}, {2, -4, 11}};
double b[N] = {6, -10, 27};
double L[N][N], D[N], x[N];
int i, j;
ldl(A, L, D, N);
solve(L, D, b, x, N);
printf("The solution is:\n");
for (i = 0; i < N; i++) {
printf("x[%d] = %f\n", i, x[i]);
}
return 0;
}
```