HDU-1969
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; }