将n个半径为a[i]的馅饼分给f+1个人，每个人分到的馅饼体积一样，求每个人能获取馅饼体积的最大值。
求二分区间满足条件的最大值。
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1e4 + 10;
const double eps = 1e-10;// 精确度
const double pi = 4*atan(1.0); // 2π
int n, f;
double a[MAXN];
bool check(double mid) {
double s = mid * mid;
int num = 0;
for (int i = 0; i < n; ++i) {
double tmp = a[i] * a[i];
num += int(tmp / s);
}
return num >= f + 1 ? true : false;
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &f);
for (int i = 0; i < n; ++i) {
scanf("%lf", &a[i]);
}
double l = 0;
double r = 10000.0;
double ans = 0;
while (fabs(r - l) > eps) {
double mid = (l + r) / 2.0;
if (check(mid)) {
ans = mid;
l = mid;
} else {
r = mid;
}
}
printf("%.4lf\n", ans * ans * pi);
}
return 0;
}