有一個橢圓方程 ax^2+by^2+cz^2+dyz+exz+fxy ,其中輸入為，a,b,c,d,e,求結果精確到小數點後8位

百度了一下，學習到了模擬退火法，寫這篇部落格進行記錄

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const double eps = 1e-8; //精度
const double r = 0.99;   //降溫速度
const int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 }; //向8個方向降溫
const int dy[] = { 1, -1, 0, 0, -1, 1, 1, -1 };
double a, b, c, d, e, f;
 
double dis(double x, double y, double z)        //求距離
{   
    return sqrt(x * x + y * y + z * z);
}
 

double getz(double x, double y)           //已知x,y,求z，二元一次方程
{ 
    double A = c, B = e * x + d * y,C = a * x * x + b * y * y + f * x * y - 1;
    double delta = B * B - 4 * A * C;
    if (delta < 0) return 1e60;
    double z1 = (-B + sqrt(delta)) / 2 / A,z2 = (-B - sqrt(delta)) / 2 / A;
    if (z1 * z1 < z2 * z2) 
    return z1;
    else 
    return z2;
}
 
double solve()          //模擬退火
{
    double step = 1;    //步長
    double x = 0, y = 0, z;
    while (step > eps) 
    {
        z = getz(x, y);
        for (int i = 0; i < 8; ++i) 
        {
            double nx = x + dx[i] * step,
                   ny = y + dy[i] * step,
                   nz = getz(nx, ny);
            if (nz > 1e30) continue;
            if (dis(nx, ny, nz) < dis(x, y, z)) 
            {
                x = nx; y = ny; z = nz;
            }
        }
        step *= r;
    }
    return dis(x, y, z);
}
 
int main() 
{
    while (scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &e, &f) != EOF) 
    {
        printf("%.8f\n", solve());
    }
    return 0;
}