我使用的是二分法。当我写如下代码时，我的第三个数据点会错误，但当我把第3行的EPS从1e-10改为1e-5时，我通过了本题。期间我一直在试图优化精度，但从来没有奏效过。最后突发奇想会不会是写数据生成器的答卷精度不够高，自己就把精度调低了，结果反而AC了。

#include <iostream>
const int MAXN(1002);
const double EPS(1e-10);
using namespace std;
int n, w[MAXN], wx[MAXN], wy[MAXN];
inline double sq(double x) { return x * x; }
inline double fisr(double x) {
	using ull = unsigned long long;
	ull b(*(ull*)&x);
	double halfx(0.5 * x);
	b = 0x5FE7000000000000ULL - (b >> 1);    // wtf?
	x = *(double*)(&b);
	x *= (1.5 - halfx * x * x);
	x *= (1.5 - halfx * x * x);
	x *= (1.5 - halfx * x * x);	// this can be removed.
	x *= (1.5 - halfx * x * x);	// this can be removed.
	return x;
}
int main() {
	double xmin(1e4), xmax(-1e4), ymin(1e4), ymax(-1e4);
	scanf("%d", &n);
	for (int i(1); i <= n; ++i) {
		scanf("%d%d%d", wx + i, wy + i, w + i);
		xmin = min(xmin, (double)wx[i]);
		xmax = max(xmax, (double)wx[i]);
		ymin = min(ymin, (double)wy[i]);
		ymax = max(ymax, (double)wy[i]);
	}
	xmin -= EPS; xmax += EPS; ymin -= EPS; ymax += EPS;
	double xmid, ymid, fx, fy;
	while (xmax - xmin > EPS || ymax - ymin > EPS) {
		xmid = (xmin + xmax) * 0.5;
		ymid = (ymin + ymax) * 0.5;
		fx = 0.0;
		fy = 0.0;
		for (int i(1); i <= n; ++ i) {
			double tmp(fisr(sq(wx[i] - xmid) + sq(wy[i] - ymid)) * w[i]);
			if (sq(wx[i] - xmid) > EPS)
				fx += (wx[i] - xmid) * tmp;
			if (sq(wy[i] - ymid) > EPS)
				fy += (wy[i] - ymid) * tmp;
		}
		if (fx > abs(fy) - EPS)
			xmin = xmid;
		if (fx < EPS - abs(fy))
			xmax = xmid;
		if (fy > abs(fx) - EPS)
			ymin = ymid;
		if (fy < EPS - abs(fx))
			ymax = ymid;
		// printf("%6lf~%6lf %6lf~%6lf\n", xmin, xmax, ymin, ymax);
	}
	printf("%.3lf %.3lf", xmid, ymid);
	return 0;
}