LA3635 - Pie

浮点型答案的二分。注意精度问题。关键代码：

while (R - L > 1e-5) {
	double M = (L + R) / 2;
	if (check(M)) L = M;
	else R = M;
}

完整代码：

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <iostream>
#include <string>
#include <map>
#include <vector>
#define MAXN 10010
using std::cin;
using std::cout;
using std::endl;
using std::map;
using std::string;
using std::vector;
using std::min;
using std::max;
int N, F;
const double PI = acos(-1.0);
double area[MAXN];
bool check(double a) {
	int sum = 0;
	for (int i = 0; i < N; ++ i) {
		sum += floor(area[i] / a);
	}
	if (sum >= F) return true;
	return false;
}
int main() {
	//freopen("input.in", "r", stdin);
	int t;
	cin >> t;
	while (t --) {
		cin >> N >> F;
		F ++;
		double Max = -1.0;
		for (int i = 0; i < N; ++ i) {
			double r;
			scanf("%lf", &r);
			area[i] = PI * r * r;
			Max = max(area[i], Max);
		}
		double L = 0, R = Max;
		while (R - L > 1e-5) {
			double M = (L + R) / 2;
			if (check(M)) L = M;
			else R = M;
		}
		printf("%.4lf\n", L);
	}
}