全选主元高斯消去算法

这是一个使用全选主元高斯消去算法来解决线性方程组的C语言实现。算法通过高斯消元逐步化简矩阵,寻找最大元素并进行行交换和列交换,最终求得解。代码中包含读取方程组、显示解的功能,并能判断解的存在性和唯一性。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

/*
    全选主元高斯消去算法,求解线性方程组 A * x = b
    By Kingwei  2005.2.27
*/

#include <stdio.h>
#include <math.h>

#define EPS        1e-11

#define MAX_DIM        20

int GaussLEquation(double x[], int dim, double A[][MAX_DIM], double b[])
{
    int maxColPos[MAX_DIM], maxRowPos;
    double maxElem, absElem, temp;
    int k, i, j, status = 1;
    
    for (k=0; k<dim; k++)
    {
        maxElem = 0.0;
        for (i=k; i<dim; i++)
            for (j=k; j<dim; j++)
            {
                absElem = fabs(A[i][j]);
                if (maxElem < absElem)
                {
                    maxElem = absElem;
                    maxRowPos = i;
                    maxColPos[k] = j;
                    }
               }
               
        if (maxElem <= EPS)
        {
            for (i=k; i<dim; i++)
                if (fabs(b[i]) > EPS)
                    return 0;
            for (i=k; i<dim; i++)
                maxColPos[i] = i;
            status = 2;
            break;
          }
          
          if (maxRowPos != k)
          {
              for (j=k; j<dim; j++)
              {
                  temp = A[k][j];
                  A[k][j] = A[maxRowPos][j];
                  A[maxRowPos][j] = temp;
                 }
                 temp = b[k];
                 b[k] = b[maxRowPos];
                 b[maxRowPos] = temp;
            }
            
            if (maxColPos[k] != k)
            {
                for (i=0; i<dim; i++)
                {
                    temp = A[i][k];
                    A[i][k] = A[i][maxColPos[k]];
                    A[i][maxColPos[k]] = temp;
                   }
              }
              
              temp = A[k][k];
              for (j=k; j<dim; j++)
                  A[k][j] /= temp;
         b[k] /= temp;
         
         for (i=k+1; i<dim; i++)
         {
             temp = A[i][k];
             for (j=k; j<dim; j++)
                 A[i][j] -= A[k][j]*temp;
            b[i] -= b[k]*temp;
           }
     }
     
     x[dim-1] = b[dim-1];
     for (i=dim-1; i>=0; i--)
     {
         temp = 0.0;
         for (j=dim-1; j>i; j--)
             temp += x[j]*A[i][j];
        x[i] = b[i]-temp;
      }
      
      for (i=dim-1; i>=0; i--)
          if (maxColPos[i] != i)
          {
              temp = x[i];
              x[i] = x[maxColPos[i]];
              x[maxColPos[i]] = temp;
            }
            
    return status;
}

int ReadLEquation(int *dim, double A[][MAX_DIM], double b[])
{
    int i, j;
    if (scanf("%d", dim) == EOF)
        return 0;
    for (i=0; i<*dim; i++)
    {
        for (j=0; j<*dim; j++)
            scanf("%lf", &A[i][j]);
        scanf("%lf", &b[i]);
    }
    return 1;
}

void DisplayRoot(double x[], int dim)
{
    int i;
    for (i=0; i<dim; i++)
        printf("%lf ", x[i]);
    printf("/n/n");
}

int main()
{
    int dim;
    double A[MAX_DIM][MAX_DIM], b[MAX_DIM], x[MAX_DIM];

    while (ReadLEquation(&dim, A, b))
    {
        switch (GaussLEquation(x, dim, A, b))
        {
            case 0:    printf("No Solution!/n/n");
                   break;
            case 1:    printf("Succeseful Solved!/n");
                DisplayRoot(x, dim);
                    break;
            case 2:    printf("Unnumberable Solutions!/n");
                   DisplayRoot(x, dim);
                       break;
            default:break;
        }
    }
    
    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值