Description:

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input:

One line with a positive integer: the number of test cases. Then for each test case:
—One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
—One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output:

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input:

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output:

25.1327
3.1416
50.2655

题目链接

直接二分结果，统计当前二分结果下是否能够分得满足需求的蛋糕。

AC代码:

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout<<#x<<"="<<x<<endl;
#define ArrayDebug(x,i) cout<<#x<<"["<<i<<"]="<<x[i]<<endl;
#define print(x) out(x);putchar('\n')
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double pi = acos(-1.0);
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) {
        return 0;
    }
    while (c != '-' && (c < '0' || c > '9')) {
        c = getchar();
    }
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') {
        ret = ret * 10 + (c - '0');
    }
    ret *= sgn;
    return 1;
}
template <class T>
inline void out(T x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) {
        out(x / 10);
    }
    putchar(x % 10 + '0');
}

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	int T;
	read(T);
	for (int Case = 1, N, F; Case <= T; ++Case) {
		read(N); read(F);
		// 加上自己
		F++;
		// 读入半径，计算面积
		vector<double> S(N);
		for (int i = 0, R; i < N; ++i) {
			read(R);
			S[i] = pi * R * R;
		}
		// 二分结果，Right=1e8会WA
		double Left = 0, Right = 1e9;
		// eps=1e-4会WA,1e-8会TLE
		while (Right - Left > eps) {
			double Mid = (Left + Right) / 2;
			int Cnt = 0;
			for (int i = 0; i < N; ++i) {
				Cnt += int(S[i] / Mid);
			}
			// 可再分
			if (Cnt >= F) {
				Left = Mid;
			}
			// 不可再分
			else {
				Right = Mid;
			}
		}
		printf("%.4lf\n", Left);
	}
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}