POJ-1129 Pie 解题报告

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：http://poj.org/problem?id=3122

       解法类型：二分法

       解题思路：v与人数的关系为单调函数，即用二分法快速查找。但要注意此题的精度问题，好像很坑。话说，样例都没有过，都AC了。。= =！

       算法实现：

//STATUS:C++_AC_32MS_172K#include<stdio.h>#include<math.h>#define PI 3.14159265358;   //注意精度const int MAXN=10010;int pie[MAXN];int main(){//freopen("in.txt","r",stdin);int t,n,f,i,tot;double mid,high,low,v;scanf("%d",&t);while(t--){low=0,high=-1;scanf("%d%d",&n,&f);f++;       //注意人数，包括自己for(i=0;i<n;i++){scanf("%d",&pie[i]);if(pie[i]>high)high=pie[i];}for(high++,mid=(low+high)/2;high-low>1e-6;mid=(low+high)/2){  //这里1e-6改成1e-5样例没有过，AC了~for(i=0,tot=0;i<n;i++){tot+=(int)floor((pie[i]*pie[i])/(mid*mid));}tot<f?high=mid:low=mid;   //二分v=mid*mid*PI;}printf("%.4lf\n",v);}return 0;}