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 ≤ 10000: the number of pies and the number of friends.
• One line with N integers ri with 1 ≤ ri ≤ 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 10−3 .
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
解题思路：
大致题意为就是公平地分披萨pie，有n个pie，找来f个朋友，加上自己那么总人数共f+1人（注意）。每个pie都是高为1的圆柱体，输入这n个pie的半径，如果要公平地把pie分给每一个人（就是所有人得到的pie尺寸一致，但是形状可以不同），而且每个人得到的那份pie必须是从同一个pie上得到的（注意这个陷阱）就是说如果有3个pie, 尺寸分别为1,2,3，如果要给每人尺寸为2的pie，那么最多分给2个人，而不是3个人因为第一个pie尺寸为1，小于2，应该被扔掉。第二个pie尺寸为2，等于2，刚好分给一个人。第三个pie尺寸为3，切出尺寸为2的一份，分给一个人，剩下的尺寸为1的就扔掉千万不要陷入 (1+2+3)/2=3人的误区，这样就变成求平均了（注意）这是一个二分法的题目，我们不断进行二分演算就可以了，还要注意的就是精度的问题。
程序代码：
#include<iostream> #include<iomanip> using namespace std; const double pi=3.14159265359; //百度搜pi就有了，我逐位提交，这是最短的pi长度，再短就WA了 //懒得测试精度的同学就把尽可能多位数的pi放进程序，肯定不会WA const double esp=1e-6; //根据题目要求的精度，为了实数二分法设定的最小精度限制值 int main(void) { int test; cin>>test; while(test--) { int n,f; //n:pie数 f:朋友数 cin>>n>>f; double* v=new double[n+1]; //每个pie的size f++; //加上自己的总人数 double maxsize=0.0; for(int i=1;i<=n;i++) { cin>>v[i]; v[i]*=v[i]; //半径平方，计算pie的体积时先不乘pi，为了提高精度和减少时间 if(maxsize<v[i]) maxsize=v[i]; } double low=0.0; //下界，每人都分不到pie double high=maxsize;//上界，每人都得到整个pie，而且那个pie为所有pie中最大的 double mid; while(high-low>esp) //还是那句，实数double的二分结束条件不同于整数int的二分结束条件 { mid=(low+high)/2; //对当前上下界折中，计算"如果按照mid的尺寸分pie，能分给多少人" int count_f=0; //根据mid尺寸能分给的人数 for(int i=1;i<=n;i++) //枚举每个pie count_f+=(int)(v[i]/mid); //第i个pie按照mid的尺寸去切，最多能分的人数（取整） //就是说如果mid尺寸为1.5，pie总尺寸为2，那么这个pie最多分给一个人 //剩下的0.5要扔掉 if(count_f < f) //当用mid尺寸分，可以分的人数小于额定人数 high=mid; //说明mid偏大，上界优化 else low=mid; //否则mid偏小，下界优化（注意'='一定要放在下界优化，否则精度会出错） } cout<<fixed<<setprecision(4)<<mid*pi<<endl; //之前的计算都只是利用半径平方去计算，最后的结果要记得乘pi delete v; } return 0; }
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double pi=acos(-1.0);
const double e=1e-6;
double a[10005];
int main()
{
int t;
cin>>t;
while(t--)
{
int n,f;
cin>>n>>f;
f++;
double v=0.0;
for(int i=0;i<n;i++)
{
cin>>a[i];
a[i]=a[i]*a[i];
if(a[i]>v)
v=a[i];
}
double l=0.0;
double r=v;
double m;
while(r-l>e)
{
m=(l+r)/2;
int s=0;
for(int j=0;j<n;j++)
s=s+a[j]/m;
if(s<f)
r=m;
else
l=m;
}
printf("%.4f\n",m*pi);
}
return 0;
}