求解线性方程组
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define ROW 100
#define COL 100

int guass(double *[ROW], double [ROW],int, int);
void exchange(double *a[ROW], int flag, int row);
int main()
{
    FILE *fp1,*fp2;
    double a[ROW][COL],b[ROW];
    int i=0,j=0,number=0,col=0,row=0;

    char ch;

    //打开文件
    if((fp1=fopen("d:\\data1.txt","r"))==NULL)
    {
        printf("File open error!\n");
        exit(0);
    }
    if((fp2=fopen("d:\\data2.txt","r"))==NULL)
    {
        printf("File open error!\n");
        exit(0);
    }
    //将线性方程的数据读入到数组中
    fseek(fp1,0,0);

    for(i=0; i<100; i++)
    {
        for(j=0; j<100; j++)
        {
            fscanf(fp1,"%lf",&a[i][j]);
        }
        fscanf(fp1,"\n");
    }
    ch=fgetc(fp1);
    while(!feof(fp1))
    {
        if (ch>='0'&&ch<='9')
            number++;
        ch=fgetc(fp1);
    }
    for(i=0; i<100; i++)
    {
        fscanf(fp2,"%lf",&b[i]);
        fscanf(fp2,"\n");
    }
    ch=fgetc(fp2);//计算系数矩阵的行数
    while(!feof(fp2))
    {
        if (ch>='0'&&ch<='9')
            row++;
        ch=fgetc(fp2);
    }
    col=number/row;//计算系数矩阵的列数


    guass(a,b,row,col+1);

    //进行回代求解

    double x[row];
    x[row-1]=b[row-1]/a[row-1][row-1]; //因矩阵化为了上三角矩阵，即可求出最后一行的解，接下来从最后一行往上一行进行回代。
    int sum=0;
    for (i=row-2; i>=0; i--)
    {
        for(j=i+1; j<col; j++)
            sum+=a[i][j]*x[j]; //累加和。
        x[i]=b[i]-sum;
        x[i]=x[i]/a[i][i];

    }
        printf("solution is :\n");
          for(i=0;i<row;i++)
          printf("%lf\n",x[i]);

    //关闭文件
    if (fclose(fp1))
    {
        printf("Can not close the file!\n");
        exit(0);
    }
    if (fclose(fp2))
    {
        printf("Can not close the file!\n");
        exit(0);
    }



    return 0;
}






void exchange(double *a[ROW], int flag, int row)
{
    int i;
    double *temp;

    for(i=flag+1; i<row; i++)
        //同一列的数依次进行比较大小
        if(fabs(*(a[flag]+flag))<fabs(*(a[i]+flag)))
        {
            //将每一列的最大值放至主对角线上
            temp = a[flag];
            a[flag] = a[i];
            a[i] = temp;
        }

    return;
}
int guass(double *a[ROW],double b[ROW],int row,int col)
{
    int temp=0,j;

    for (int i=0; i<row-1; i++)
    {
        exchange(a,i,row);
        for(j=i+1; j<row; j++)
            temp=-*(a[j]+i)/ *(a[i]+i);
        b[j]+=b[i]*temp;
        for(int k=i; k<col; k++)
           *(a[j]+k) += temp * *(a[i]+k);

    }
    return;
}

[此贴子已经被作者于2019-6-10 22:23编辑过]