LA3635 Pie (二分判定)

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 com-

plaining. 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

10000: the number of pies and the number

of friends.

One line with

N

integers

r

i

with 1

r

i

10000: 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 oating point number with an absolute

error of at most 10

这个判断能否达到最小的问题，是一个比较裸的二分判定问题，我们二分孩子可以得到的面积，就可以得答案了。

以后碰到这种已经知道了上下界，然后让你求最小解的问题，应当转化为二分来解决。

#include<cstdio>
#include<cmath>

const int maxn=;
const double PI=acos(-);
const double VALUE=;
double a[maxn];
int n,m;

int Judge(double area){
    int t=;
    for(int i=;i<n;++i){
        double fl=a[i];
        while(fl>=area){
            fl-=area;
            ++t;
        }
    }
    return t;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        double all=;
        for(int i=;i<n;++i){
            scanf("%lf",&a[i]);
            a[i]=a[i]*a[i]*PI;
            all+=a[i];
        }
        ++m;
        double left=,right=all,mid;
        while(left<=right){
            mid=(left+right)/;
            int nums=Judge(mid);
            if(nums>=m)left=mid+VALUE;
            else right=mid-VALUE;
        }
        printf("%.4f\n",left);
    }
    return ;
}