Question

用C语言写一个高斯消元法解方程组的程序

不要用列主消元的啊
就是化成上三角阵直接求的那种

越快越好哦 谢谢各位了！

Answer 1

我们以方程组 2x1 + 6x2 - x3 = -12
5x1 - x2 +2x3 = 29
-3x1 - 4x2 + x3 = 5
为例 来说明楼主自己把方程组化为矩阵形式。以下为源代码 。
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <math.h>

int GS(int,double**,double *,double);
double **TwoArrayAlloc(int,int);
void TwoArrayFree(double **);

int main(void)
{
int i,n;
double ep,**a,*b;
n = 3;
ep = 1e-4;

a = TwoArrayAlloc(n,n);
b = (double *)calloc(n,sizeof(double));
if(b == NULL)
{
printf("memory get error\n");
exit(1);
}
a[0][0]= 2; a[0][1]= 6; a[0][2]= -1;
a[1][0]= 5; a[1][1]=-1; a[1][2]= 2;
a[2][0]=-3; a[2][1]=-4; a[2][2]= 1;
b[0] = -12; b[1] = 29; b[2] = 5;
if(!GS(n,a,b,ep))
{
printf("can't solve it with GS elimination\n");
exit(0);
}
printf("The solution of equations is as follows：\n");

for(i=0;i<3;i++)
{
printf("x%d = %.2f\n",i,b[i]);
}
TwoArrayFree(a);
free(b);

return 0;
}

int GS(n,a,b,ep)
int n;
double **a;
double *b;
double ep;
{
int i,j,k,l;
double t;

for(k=1;k<=n;k++)
{
for(l=k;l<=n;l++)

if(fabs(a[l-1][k-1])>ep)
break;
else if(l==n)
return(0);
if(l!=k)
{
for(j=k;j<=n;j++)
{
t = a[k-1][j-1];
a[k-1][j-1] =a[l-1][j-1];
a[l-1][j-1] =t;
}
t=b[k-1];
b[k-1]=b[l-1];
b[l-1]=t;
}
t=1/a[k-1][k-1];
for(j=k+1;j<=n;j++)
a[k-1][j-1]=t*a[k-1][j-1];
b[k-1]*=t;
for(i=k+1;i<=n;i++)
{
for(j=k+1;j<=n;j++)
a[i-1][j-1]-=a[i-1][k-1]*a[k-1][j-1];
b[i-1]-=a[i-1][k-1]*b[k-1];
}
}

for(i=n-1;i>=1;i--)
for(j=i+1;j<=n;j++)
b[i-1]-=a[i-1][j-1]*b[j-1];
return(1);
}

double **TwoArrayAlloc(int r,int c)
{
double *x,**y;
int n;
x=(double *)calloc(r*c,sizeof(double));
y=(double **)calloc(r,sizeof(double*));

for(n=0;n<=r-1;++n)
{
y[n]=&x[c*n];
}
return y ;
}

void TwoArrayFree(double **x)
{
free(x[0]);
free(x);
}

追问

能不能调整成输入任意一个矩阵的形式呢？
谢了！

追答

无非要多加个矩阵维数和存储数组的维数的输入。你在我的基础上作点修改啊。这又不算难

Answer 2