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
题目大意：给你N个蛋糕，k+1个人，问每个人能获得的最大体积的蛋糕是多少。
解题思路：m=选出最大的蛋糕体积/k+1人，p=全部蛋糕的体积和/k+1；
每次用m+p/2，来除以每个蛋糕的体积，如果人数相等或大于，这个体积就是最大的。
代码：
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<cmath>
5 #include<cstring>
6 using namespace std;
7 const double pi=acos(-1.0);
8 const int maxn=10000+10;
9 int main()
10 {
11 int T,N,F,i,c;
12 double l,r,mid,a[maxn],sum;
13 scanf("%d",&T);
14 while(T--)
15 {
16 memset(a,0,sizeof(a));
17 scanf("%d%d",&N,&F);
18 F++;
19 sum=0;
20 for(i=0;i<N;i++)
21 {
22 scanf("%d",&c);
23 a[i]=pi*c*c;
24 sum+=a[i];
25 }
26 sort(a,a+N);
27 l=a[N-1]/F;
28 r=sum/F;
29 while(r-l>1e-6)
30 {
31 int count=0;
32 mid=(l+r)/2;
33 for(i=0;i<N;i++)
34 count+=(int)floor(a[i]/mid);
35 if(count>=F)
36 l=mid;
37 else r=mid;
38 }
39 printf("%.4f\n",l);
40 }
41 return 0;
42 }