求球面两点最短距离

Problem:
给定球的半径，两点的经纬度，求这两点间的最短距离。
Solution:
球面两点间距离公式：
r*acos(cos(wa)*cos(wb)*cos(jb-ja) + sin(wa)*sin(wb))
r代表半径，wa是a点的纬度，wb是b点的纬度，ja是a点的经度，jb是b点的经度，去北纬为正，东经为正。
note:
cout输入输出有一定的技巧需要掌握。

#include
#include
#include 

using namespace std;

const double PI = 3.141592653589;

//degree to its radian number
double convert(double degree) {
    return degree*PI/180;
}

//parameters are radian number
double calculate(double r, double wa, double ja, double wb, double jb) {
    return r*acos(cos(wa)*cos(wb)*cos(jb-ja) + sin(wa)*sin(wb));
}

int main() {
    double r, waa, wbb, jaa, jbb;
    char c;

    while(cin >> r) {
        if(r == 0)
            break;
        cin >> c >> waa;
        if(c == 'S')
            waa = -waa;
        cin >> c >> jaa;
        if(c == 'W')
            jaa = -jaa;
        cin >> c >> wbb;
        if(c == 'S')
            wbb = -wbb;
        cin >> c >> jbb;
        if(c == 'W')
            jbb = -jbb;
        cout << fixed << setprecision(2) << calculate(r, convert(waa), convert(jaa), convert(wbb), convert(jbb)) << endl;
    }

    return 0;
}