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2015-08-05 09:20 309 查看

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 109 Accepted Submission(s): 52

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).

[align=left]Sample Input[/align]

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

[align=left]Sample Output[/align]

25.1327
3.1416
50.2655

题意：

一个人要办生日part，有f个馅饼，n个朋友要
9957
来参加他的生日part，我们要做的是让着f个馅饼平均分给他的朋友，

形状可以不一样，但是大小要一样，也就是说馅饼的体积要一样，这馅饼的高都为1。这个简单，但是题目要求不能分

从一块一块的凑出来的馅饼。

思路：
用二分搜索，首先算出馅饼的总体积，然后再除以总人数，要注意总人数要加1，因为生日的主人也包括在内，总体积就是
大家能够的到的最大馅饼的体积，但是不能从一块到另一块凑出来，所以要进行二分搜索，mid=(l+r)/2,当达到题意是就退出
循环，输出mid.这里PI的精度要高一些，用acos(-1.0)。
代码：
#include<cstdio>
#include<cmath>
#define PI acos(-1);
double v[10001];
int main()
{
int T,n,f,ri;
double l,r,mid,sum;
scanf("%d",&T);
while(T--)
{

r=l=0;
scanf("%d%d",&n,&f);
f++;
for(int i=0;i<n;i++)
{
scanf("%d",&ri);
v[i]=ri*ri*PI;
r+=v[i];

}
r/=f;
while(r-l>=1e-6)
{

sum=0;
mid=(l+r)/2.0;
for(int i=0;i<n;i++)
sum+=(int)(v[i]/mid);
if(sum>=f)
l=mid;
else r=mid;
}
printf("%.4lf\n",mid);

}
return 0;
}

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