hd1969 Pie

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6612    Accepted Submission(s): 2506

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

33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2

 

Sample Output

25.13273.141650.2655

 

恩，题目大意就是说，有n块半径不同的蛋糕，高度为1，想要平均分配给m个人，问每个人能得到的最大体积的蛋糕，二分查找最适合的半径的平方。

恩，还是不怎么会做二分的题。。。

#include<cstdio>#include<cmath>#include<cstring>#define PI acos(-1.0)int ri[10010];int t,n,m;int judge(double r){int cnt=0;for(int i=0;i<n;++i){cnt+=(int)(ri[i]*1.0/r);}return cnt>m?1:0;//不造为什么一开始我写的cnt<m?1:0;然后下面的也相应改了，但是怎么都不对，后来一步一步调试才发现问题，然后换成现在这样就过了，还是不很理解为什么，感觉都一样啊}int main(){double left,right,mid;scanf("%d",&t);while(t--){double maxi=0;memset(ri,0,sizeof(ri));scanf("%d %d",&n,&m);for(int i=0;i<n;++i){scanf("%d",&ri[i]);ri[i]=ri[i]*ri[i];if(maxi<ri[i])maxi=ri[i];}left=0;right=maxi;while(right-left>1e-7){mid=(left+right)/2;if(judge(mid)) left=mid;else right=mid;}printf("%.4lf\n",PI*mid);}return 0;}