c语言求线性方程组组的解

以下是三种C语言求解线性方程组的方法：

1. 一般的高斯消元解法

#include <stdio.h>
#define MAX 100

void gauss(double a[MAX][MAX], double b[MAX], int n){
    int i,j,k;
    double factor, sum, x[MAX];
    for(i=0;i<n;i++){
        for(j=i+1;j<n;j++){
            factor = a[j][i]/a[i][i];
            for(k=i;k<n;k++){
                a[j][k] -= factor*a[i][k];
            }
            b[j] -= factor*b[i];
        }
    }
    for(i=n-1;i>=0;i--){
        sum = 0;
        for(j=i+1;j<n;j++){
            sum += a[i][j]*x[j];
        }
        x[i] = (b[i]-sum)/a[i][i];
    }
    printf("The solution is:\n");
    for(i=0;i<n;i++){
        printf("x[%d] = %lf\n", i, x[i]);
    }
}

2. Cholesky法求正定对称矩阵的线性方程组

#include <stdio.h>
#include <math.h>
#define MAX 100

void cholesky(double a[MAX][MAX], double b[MAX], int n){
    int i,j,k;
    double l[MAX][MAX], y[MAX], x[MAX], sum;
    for(i=0;i<n;i++){
        for(j=0;j<=i;j++){
            sum = 0;
            if(j==i){
                for(k=0;k<j;k++){
                    sum += l[j][k]*l[j][k];
                }
                l[j][j] = sqrt(a[j][j]-sum);
            }
            else{
                for(k=0;k<j;k++){
                    sum += l[i][k]*l[j][k];
                }
                l[i][j] = (a[i][j]-sum)/l[j][j];
            }
        }
    }
    y[0] = b[0]/l[0][0];
    for(i=1;i<n;i++){
        sum = 0;
        for(j=0;j<i;j++){
            sum += l[i][j]*y[j];
        }
        y[i] = (b[i]-sum)/l[i][i];
    }
    x[n-1] = y[n-1]/l[n-1][n-1];
    for(i=n-2;i>=0;i--){
        sum = 0;
        for(j=i+1;j<n;j++){
            sum += l[j][i]*x[j];
        }
        x[i] = (y[i]-sum)/l[i][i];
    }
    printf("The solution is:\n");
    for(i=0;i<n;i++){
        printf("x[%d] = %lf\n", i, x[i]);
    }
}

3. 追赶法解线性方程组

#include <stdio.h>
#define MAX 100

void chase(double a[MAX], double b[MAX], double c[MAX], double d[MAX], double x[MAX], int n){
    int i;
    double p[MAX], q[MAX];
    p[0] = -c[0]/b[0];
    q[0] = d[0]/b[0];
    for(i=1;i<n;i++){
        p[i] = -c[i]/(a[i]*p[i-1]+b[i]);
        q[i] = (d[i]-a[i]*q[i-1])/(a[i]*p[i-1]+b[i]);
    }
    x[n-1] = q[n-1];
    for(i=n-2;i>=0;i--){
        x[i] = p[i]*x[i+1]+q[i];
    }
    printf("The solution is:\n");
    for(i=0;i<n;i++){
        printf("x[%d] = %lf\n", i, x[i]);
    }
}