Pie (二分)

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1105    Accepted Submission(s): 365

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

思路 ：

            一道典型的浮点二分，分饼时不可以是拼凑的，所以从max 和 0之间二分。

            二分重要的一点在于检测条件，这里，对于每个mid 让每个饼除 mid 累加计算 num 是否够分，继而将 num 作为二分检测条件。

            注意： 浮点二分要设置精度作为比较量度（dat），二分时 l（r） = mid 而不能 = mid+1，跨度为一，对浮点太奢侈了。

#include <iostream>#include <cstdio>#include <queue>#include <algorithm>#include <cmath>using namespace std;const double PI = acos(-1.0);const int max_ = 10010;const double dat = 0.00001;//设置精度,适当最好，防止超时。double a[max_];int m,n;int check(double mid){    int tmp = 0;    for(int i = 0; i < m; i++)        tmp += (int)(a[i]/mid);    return tmp;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        double small = 0; //二分下界        double big = 0;//保留最大的饼的面积,作为二分上界        scanf("%d%d",&m,&n);        for(int i = 0; i < m; i++)        {            scanf("%lf",&a[i]);            a[i] = PI*a[i]*a[i];//存面积            big = max(big,a[i]);        }        while(big - small > dat)        {            double mid = (small + big) / 2;            int tmp = check(mid);            if(tmp > n)                small = mid;            else                big = mid;        }        printf("%.4f\n", big);    }    return 0;}