一定有解的高斯消元

#include
using namespace std;
const int N = 1005;
int n;
double a[N][N];
signed main()
{
    scanf("%d", &n);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n+1; j++)
            scanf("%lf", a[i]+j);
    for(int i=1; i<=n; i++)
    {
        int p = i;
        for(int j=i+1; j<=n; j++)
            if(fabs(a[j][i])>fabs(a[p][i])) p=j;
        swap(a[i], a[p]);
        for(int j=1; j<=n; j++)
        {
            if(j==i) continue;
            double r = a[j][i] / a[i][i];
            for(int k=i; k<=n+1; k++)
                a[j][k] -= a[i][k] * r;
        }
    }
    for(int i=1; i<=n; i++)
        printf("%d ", (int) round(a[i][n+1] / a[i][i]));
}