HDU 1173 采矿 计算几何

思路:(x1,y1)(x_1,y_1)距离（x2,y2）（x_2,y_2）是（x2−x1）+(y2−y1)（x_2-x_1）+(y_2-y_1);所以答案(x,y)(x,y)到所有点的距离为∑n1(xi−x)+∑n1(yi−y)\sum_{1}^{n}(x_i-x)+\sum_{1}^{n}(y_i-y),于是就可以单独求出,x,y的值了。

/*********************************************
Problem : HDU 1173
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)
#define rrep(i,a,b) for(int i = b ; i >= a ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

double X[MAXN],Y[MAXN];

void input() {
rep(i,1,n) scanf("%lf %lf",&X[i],&Y[i]);
}

void solve() {
sort(X+1,X+n+1);
sort(Y+1,Y+n+1);
double ansX , ansY;
if(n & 1) {ansX = X[(n+1)/2] ; ansY = Y[(n+1)/2] ;}
else {ansX = (X[n/2] + X[n/2+1]) / 2; ansY = (Y[n/2] + Y[n/2+1]) / 2 ;}
printf("%.2lf %.2lf\n",ansX,ansY);
}

int main(void) {
//freopen("a.in","r",stdin);
//scanf("%d",&T);while(T--) {
//while(~scanf("%d %d",&n,&m)) {
while(scanf("%d",&n),n) {
input();
solve();
}
return 0;
}