高斯消元

老早了解了原理，一直没写过

每次消去一个系数，从系数大的开始消，当出现方程提前为0的情况说明无解或者有无穷多解

#include <bits/stdc++.h>
using namespace std;
int n;
#define N 150
const double eps=1e-7;
double a[N][N];
bool Gauss()
{
  int r;
  double f;
  for (int i=0;i<n;i++)
  {
     r=i;
     for (int j=i+1;j<n;j++)
       if (fabs(a[j][i])>fabs(a[r][i])) r=j;
     if (fabs(a[r][i])<eps) return (false);
     if (r!=i)
       for (int j=0;j<=n;j++) swap(a[r][j],a[i][j]);
     for (int k=i+1;k<n;k++)
     {
       f=a[k][i]/a[i][i];
       for (int j=i;j<=n;j++) a[k][j]-=f*a[i][j];
     }
  }
  for (int i=n-1;i>=0;i--)
  {
    for (int j=i+1;j<n;j++) a[i][n]-=a[j][n]*a[i][j];
    a[i][n]/=a[i][i];
  }
  return true;
}
int main()
{
  freopen("noip.in","r",stdin);
  freopen("noip.out","w",stdout);
  std::ios::sync_with_stdio(false);
  cin>>n;
  int x;
  for (int i=0;i<n;i++)
    for (int j=0;j<=n;j++)
    {
      cin>>x;
      a[i][j]=x;
    }
  if (!Gauss())
  {
    cout<<"No Solution"<<endl;
    return 0;
  }
  for (int i=0;i<n;i++) printf("%.2f\n",a[i][n]);
  return 0;
}

 高斯约当消元法 常熟略大但省略回带过程

#include <bits/stdc++.h>
using namespace std;
#define N 150
const double eps=1e-7;
int n;
double a[N][N];
bool Gauss()
{
  int now; double f;
  for (int i=0;i<n;i++)
  {
    now=i;
    for (int j=i;j<n;j++)
      if (fabs(a[now][i])<fabs(a[j][i])) now=j;
    if (fabs(a[now][i])<eps) return(0);
    if (now!=i)
      for (int j=0;j<=n;j++) swap(a[now][j],a[i][j]);
    for (int j=0;j<n;j++)
      if (j!=i&&fabs(a[j][i])>eps)
      {
        f=a[j][i]/a[i][i];
        for (int k=0;k<=n;k++)
          a[j][k]-=f*a[i][k];
      }
  }
  for (int i=0;i<n;i++) a[i][n]/=a[i][i];
  return 1;
}
int main()
{
  freopen("noip.in","r",stdin);
  freopen("noip.out","w",stdout);
  std::ios::sync_with_stdio(false);
  cin>>n;
  int x;
  for (int i=0;i<n;i++)
    for (int j=0;j<=n;j++)
    {
      cin>>x;
      a[i][j]=x;
    }
  if (!Gauss())
  {
    cout<<"No Solution"<<endl;
    return 0;
  }
  for (int i=0;i<n;i++) printf("%.2f\n",a[i][n]);
  return 0; 
}