编程之美初赛第三题

三分求解 ~  

也可以求导后求导数值为0

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

using namespace std;

struct point
{
    double x, y;
};

point a[100010];
int n;

double getdis(double d)
{
    double sum = 0;
    for(int i = 1; i <= n; ++ i)
    {
        sum += sqrt((a[i].x-d)*(a[i].x-d)+a[i].y*a[i].y);
    }
    return sum;
}
int main()
{
    int t, ca = 0;
    scanf("%d", &t);
    while(t --)
    {
        double _min, _max;
        scanf("%d", &n);
        scanf("%lf%lf", &a[1].x, &a[1].y);
        _min = min(a[1].x, a[1].y);
        _max = max(a[1].x, a[1].y);
        for(int i = 2; i <= n; ++ i)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
            _min = min(_min, min(a[i].x, a[i].y));
            _max = max(_max, max(a[i].x, a[i].y));
        }
        while(_max-_min > 1e-8)
        {
            double mid = (_min+_max)/2;
            double midl = (mid+_max)/2;
            double a = getdis(mid), b = getdis(midl);
            if(b < a) _min = mid;
            else _max = midl;
        }
        printf("Case %d: %.6lf\n", ++ca, _min);
    }
    return 0;
}