[poj3122]pie——简单二分

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：给你几块饼的半径，将尽量大的完整的一块饼分给每一个朋友，求分到的饼的体积最大是多少。

分析：直接二分答案即可，需要注意的是分到的饼必须是一个完整的而不是多块小的拼凑而成，还有π的值精度要尽量高。

另外写这道题遇到一个十分诡异的情况，有的代码c++能过，G++就莫名其妙的wa了，orz........

ac代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define val(x) acos(-1.0)*x*x
using namespace std;
const int N = 100000;
int pie[N];
int n;
bool check(double mid,int f)
{
	int cont = 0;
	for(int i = 1;i <= n;i++)
	{
		cont += (int)(val(pie[i])/mid);
	}
	if(cont >= f)
		return true;
	return false;
}
int main()
{
    int t, f;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&f);
        int maxn = -1, minn = N;
        double sum = 0;
        for(int i = 1;i <= n; i++)
        {
            scanf("%d",&pie[i]);
            maxn = max(pie[i],maxn);
            minn = min(minn,pie[i]);
            sum += val(pie[i]);
        }
        double r = val(maxn), l = 0, mid;
        double minx = val(minn);
        while(fabs(r-l) > 1e-5)
        {
            mid = (r + l)/2.0;
            if(check(mid,f+1))  l = mid;
            else                r = mid;
        }
       printf("%.4f\n",r);
    }
    return 0;
}