double hypot(double x, double y) ...{ return sqrt(x * x + y * y);}double distance(double wd1, double jd1, double wd2, double jd2) ...{// 根据经纬度坐标计算实际距离 double x, y, out; double PI = 3.1415926535898; double R = 6.371229 * 1e6; x = (jd2 - jd1) * PI * R * cos( ( (wd1 + wd2) / 2) * PI / 180) / 180; y = (wd2 - wd1) * PI * R / 180; out = hypot(x, y); return out; }