hdoj 1969 Pie 【二分】

最新推荐文章于 2019-07-29 21:47:28 发布

原创最新推荐文章于 2019-07-29 21:47:28 发布 · 517 阅读

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本文探讨如何在有限的饼资源下，合理分配给不同数量的朋友，确保每个人获得的饼大小一致，最大化每个人能获得的饼量。通过数学方法解决实际问题，涉及圆柱体体积计算与优化分配策略。

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Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6253 Accepted Submission(s): 2368

Problem 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

题意：有F+1个人和N个饼，现在分这些饼，要求每个饼的体积一样（每个饼都是高为1的柱体，所以说求体积就是求底面积）。问你能分的最大体积（就是底面积）。

无语，太想省事了，直接右区间开到20000000，以为就OK了，没想到WA到死。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define eps 1e-6
#define MAXN 10000+10
#define PI acos(-1.0)
using namespace std;
double R[MAXN];
int N, F;
bool judge(double mid)
{
    int cnt = 0;
    for(int i = 0; i < N; i++)
        cnt += (int) (R[i] / mid);
    return cnt >= F+1;//自己也要。。。
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &N, &F);
        double r;
        for(int i = 0; i < N; i++)
        {
            scanf("%lf", &r);
            R[i] = PI * r * r;
            //sum += R[i];
        }
        double left = 0, right = 200000000000.0, mid;
        while(right - left >= eps)
        {
            mid = (left + right) / 2;
            if(judge(mid))
                left = mid;
            else
                right = mid;
        }
        printf("%.4lf\n", left);
    }
    return 0;
}