Pie
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 14536 | Accepted: 4979 | Special Judge | ||
Description
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
题目大意：国人很多事情都追求公平，分饼也是如此，现在这里有n个饼，每一个饼都可以看做一个圆柱体，高都是1，但是半径不同,
每一个人都可以分到某个饼的一部分，但是只能要一部分，而不能要好几块饼，最终结果必须保证每个人分到的饼的体积（面积）相等，
问你每个人能够获得的饼的最大面积是多少。
思路分析：首先数据量很大，如果用暴力枚举肯定会超时，很显然我们应该采用二分逼近的方法来确定答案，但是在实际操作的时候还是
出了一些问题，首先，二分精度不能够太高，1e-5就可以，精度太高会超时，其次，关于π，如果写做3.1415926会被精度卡掉，将pi
定义为acos(-1.0)就可以了。
代码：
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<iostream>
#include<algorithm>
#include <cmath>
#define LL long long
using namespace std;
const int maxn=10000+100;
const double pi=acos(-1.0);
double a[maxn];
int n,f;
double s(double r)
{
return pi*r*r;
}
bool check(double x)
{
int t=0;
for(int i=0;i<n;i++)
{
double p=s(a[i]);
while(p>=x)
{
p-=x;
t++;
if(t>=f+1) return true;
}
}
return false;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
double sum=0.0;
scanf("%d%d",&n,&f);
for(int i=0;i<n;i++)
{
scanf("%lf",&a[i]);
sum+=s(a[i]);
}
sort(a,a+n);
double l=0.0,r=sum/(f+1)*1.0;
double ans=0;
while(l+0.000001<=r)
{
double mid=(l+r)/2;
if(check(mid)) ans=mid,l=mid;
else r=mid;
}
printf("%.4lf\n",ans);
}
return 0;
}