Pie
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10629 | Accepted: 3744 | Special Judge | ||
Description
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
Source
还是附上中文题目吧
13 派（NWERC 2006, LA 3635）
有F+1个人来分N个圆形派（半径不同），每个人得到的必须是一整块派，而不是几块拼在一起，且面积要相同。求每个人最多能得到多大面积的派（不必是圆形）。
【输入格式】
输入的第一行为数据组数T。每组数据的第一行为两个整数N和F（1≤N，F≤10 000）；第二行为N个整数ri（1≤ri≤10 000），即各个派的半径。
【输出格式】
对于每组数据，输出每人得到的派的面积的最大值，精确到10-3。
思路：
显然二分答案 没什么好解释的了
附白书解释：
这个问题并不是“最小值最大”问题，但仍然可以采用二分答案方法，把问题转化为“是否可以让每人得到一块面积为x的派”。这样的转化相当于多了一个条件，然后求解目标变成了“看看这些条件是否相互矛盾”。
会有怎样的矛盾呢？只有一种矛盾：x太大，满足不了所有的F+1个人。这样，我们只需要算算一共可以切多少份面积为x的派，然后看看这个数目够不够F+1即可。因为派是不可以拼起来的，所以一个半径为r的派只能切出[πr2/x]个派（其他部分就浪费了），把所有圆形派能切出的份数加起来即可。代码如下。
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double PI=acos(-1);
const int maxn=10000+5;
int N,F,r;
double radius[maxn];
double ans;
double squre; bool ok(double area)
{
int sum = 0;
for(int i = 1; i <= N; i++) sum += floor(radius[i] / area);
return sum >= F;
} int getans()
{
double i=0,j=squre,m;
while(j-i>1e-6)
{
m=(j+i)/2;
if(ok(m)) i=m;
else j=m;
}
ans=i;
return 0;
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
squre=-1;
scanf("%d%d",&N,&F);
F++;
for(int i=1;i<=N;i++)
{
scanf("%d",&r);
radius[i]=r*r*PI;
squre=max(squre,radius[i]);
}
getans();
printf("%.4lf\n",ans);
}
return 0;
}