分析：计算几何、大地坐标系。利用公式可直接解得两点的空间距离：
d = r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)))
推导过程如下：
如图，C，D为已知两点则有如下推导：
AB = r*cos(lat1)；DE = r*cos(lat2)；BE = r*sin(lat1) + r*sin(lat2)；
AD*AD = BE*BE + (AB-DE)*(AB-DE) = 2*r*r - 2*r*r*sin(lat1)*sin(lat2) - 2*r*r*cos(lat1)*cos(lat2)；
AC*AC = 2*AB*AB - 2*AB*AB*cos(lon1-lon2) = 2*r*r*cos(lat1)*cos(lat1)*(1-cos(lon1-lon2))；
DF*DF = 2*DE*DE - 2*DE*DE*cos(lon1-lon2) = 2*r*r*cos(lat2)*cos(lat2)*(1-cos(lon1-lon2))；
AC*DF = 2*r*r*cos(lat1)*cos(lat2)*(1-cos(lon1-lon2))；
由托勒密定理有 AC*DF + AD*AD = CD*CD 整理有：
CD = r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)))；
注意：输出最近的整数，%.0lf即可。
#include#include #include int main() { double r = 6371009; double p = acos(-1.0); int n; double l1,d1,l2,d2; while ( scanf("%d",&n) != EOF ) while ( n -- ) { scanf("%lf%lf%lf%lf",&l1,&d1,&l2,&d2); l1 *= p/180.0; l2 *= p/180.0; d1 *= p/180.0; d2 *= p/180.0; double d = r*sqrt(2-2*(cos(l1)*cos(l2)*cos(d1-d2)+sin(l1)*sin(l2))); printf("%.0lf\n",2*asin(d/(2*r))*r-d); } return 0; }