来源:http://blog.csdn.net/liminlu0314/article/details/8553926
此公式得解，用 acos 函数听说会有点慢
//====================================================================||
// ||
// ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 1005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
int n;
double pi=acos(-1.0);
struct pt{
double lon,lat;
double an,at;
}p\[M\];
double dis(pt a,pt b){
return acos(cos(a.lat)\*cos(b.lat)\*cos(b.lon-a.lon)+sin(a.lat)\*sin(b.lat));
}
void solve(){
double ans=oo;
int id=-1;
for(int i=0;i<n;i++){
double mx=0;
for(int j=0;j<n;j++){
if(i==j)continue;
double tmp;
if(xdy(tmp=dis(p\[i\],p\[j\]),mx)){
mx=tmp;
}
}
if(xcdy(mx,ans)){
ans=mx;
id=i;
}
}
printf("%.2f %.2f\\n",p\[id\].at,p\[id\].an);
}
double tr(double x){
return x\*pi/180;
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d",&n)){
for(int i=0;i<n;i++){
scanf("%lf%lf",&p\[i\].at,&p\[i\].an);
p\[i\].lat=tr(p\[i\].at);
p\[i\].lon=tr(p\[i\].an);
}
solve();
}
}