存份模板（模板洛咕$P3389$）

1 #include 
      
        2  3 int n; 4  5 double a[200][200], b[200]; 6  7 int main() 8 { 9     scanf("%d", &n);10     for (int i = 1; i <= n; i++)11     {12         for (int j = 1; j <= n; j++)13             std::cin >> a[i][j];14         std::cin >> b[i];15     } 16     17     for (int i = 1; i <= n; i++)18     {19         for (int j = i; j <= n; j++)20         {21             if (fabs(a[j][i]) > 1e-8)22             {23                 for (int k = 1; k <= n; k++)24                     std::swap(a[i][k], a[j][k]);25                 std::swap(b[i], b[j]);26             }27         }28         if(fabs(a[i][i]) < 1e-8)29         {30             puts("No Solution");31             return 0;32         }33         for (int j = 1; j <= n; j++)34         {35             if (i == j)36                 continue;37             double rate = a[j][i] / a[i][i];38             for (int k = i; k <= n; k++)39                 a[j][k] -= a[i][k] * rate;40             b[j] -= b[i] * rate;41         }42     }43     for (int i = 1; i <= n; i++)44         printf("%.2f\n", b[i] / a[i][i]);45     return 0;46 }