题意

Sol

欲哭无泪啊qwq。。。。昨晚一定是智息了qwq

说一个和标算不一样做法吧。。

显然\(x\)轴是可以三分的，半径是可以二分的。

恭喜你获得了一个TLE的做法。。

然后第二维的二分是没有必要的，直接拿圆的标准方程推一下取个最大值就行了。。。。。昨晚没想到qwq给数学老师丢脸了。。

#include
    
     #include
     
      #include
      
        #define double long double using namespace std;const double eps = 1e-7, INF = 1e18;const int MAXN = 1e5 + 10;inline int read() {    char c = getchar(); int x = 0, f = 1;    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}    while(c >= '0' && c <= '9') x = x * 10 + c - '0',c = getchar();    return x * f;}int N, up, down;double max(double a, double b) {return a > b ? a : b;}double min(double a, double b) {return a < b ? a : b;}struct Node {    double x, y;}a[MAXN];int check(int x, int y) {    if(x < 0 && y > 0) return 1;    else return 0;}double mxr;double sqr(double x) {    return x * x;}double f(double x) {    double mx = 0;    for(int i = 1; i <= N; i++)         mx = max(mx, fabs((sqr(a[i].x - x) + sqr(a[i].y)) / (2.0 * a[i].y)));    return mx;}      int main() {    N = read();    double L = INF, R = -INF;    for(int i = 1; i <= N; i++) {        a[i].x = read(), a[i].y = read();        up = min(up, a[i].y);        mxr = max(a[i].y, mxr);        L = min(a[i].x, L);         R = max(a[i].x, R);    }    if(check(up, mxr)) {puts("-1"); return 0;}    mxr = INF;    if(up < 0) for(int i = 1; i <= N; i++) a[i].y = -a[i].y;    int tim = 100;    while(tim--) {        double x = (2 * L + R) / 3, y = (L + 2 * R) / 3;        f(x) < f(y) ? R = y : L = x;      //  printf("%Lf %Lf\n", f(x), f(y));    }    printf("%.10Lf", f(L));    return 0;   }