课程练习二-1003-pie

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

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

 

题意：

—有f+1个人分n块披萨，每个人要求分得的面积一样，且披萨只能被切开而不能重新组合，求每个人能分到的最大面积v。

注意：F个朋友+自己

思路：

每个人要得到一张完整的饼，而不能多个组合。对最大的饼（high）与0（low）进行二分,判断mid是否sum(size[i])/mid是否大于等于F+1；

AC代码：

#include
#include
#include
#include
#include
#include
using namespace std;


bool judge(double mid,int n,int f,double size[])
{
int  p = 0;
for (int i = 0; i < n; ++i)
{
p += int(size[i] / mid);
}
return p >= f+1;
}


int main()
{
int n;
cin >> n;
while (n--)
{
int N, F;
double pi = acos(-1);
cin >> N >> F;
double size[10001];
memset(size, 0, sizeof(size));
int a = 0;
for (int i = 0; i < N; i++)
{
cin >> a;
size[i] = a*a*pi;  //把每个饼的面积存入数组，本来是将半径存入数组最后进行pi*r*r的，结果总是通不过测试数据。
}
double high = size[0];
for (int i = 0; i < N; i++)
{
if (size[i]>high)
{
high = size[i];
}
}
double mid;
double low = 0;
double ans = 0;
//cout << high << endl;
while ((high - low) > 0.00001)
{
mid = (high + low) / 2;
//cout << judge(mid, N, F, size) << endl;
if (judge(mid, N, F, size))
{
low = mid;
ans = mid;
}
else {
high = mid;
}
}
//cout << ans << endl;
double jg = ans;
cout << setprecision(4) << fixed < }
system("pause");
return 0;
}