虽然但是
这边推荐 $\texttt{Gauss-Jordan}$ 消元(
直接消成对角矩阵(
```cpp
#define abs fabs //骗过上帝
inline bool solve()
{
ri sta;
for(ri i=1;i<=n;i++)
{
sta=i;
for(ri j=i+1;j<=n;j++)
if(abs(a[j][i])>abs(a[sta][i]))
sta=j;
if(sta!=i)
for(ri j=1;j<=n+1;j++)
swap(a[sta][j],a[i][j]);
if(a[i][i]==0)
return true;
for(ri j=1;j<=n;j++)
{
if(j==i) continue;
for(ri k=i+1;k<=n+1;k++)
a[j][k]-=a[i][k]*a[j][i]/a[i][i];
}
}
return false;
}
```
by Neutralized @ 2022-02-16 16:23:59
第22行倒过来循环
by sonderl @ 2022-02-16 16:26:53
@[sonderl](/user/110754) tql!过了,已关注。谢谢!
by Wsyflying2022 @ 2022-02-16 17:24:15
@[Neutralized](/user/538609) 你讲的我听不懂(QAQ)
by Wsyflying2022 @ 2022-02-16 17:24:54