POJ 3122 Pie 二分答案

Pie Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9653 Accepted: 3478 Special Judge

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

Source

Northwestern Europe 2006

題解

題意是將，一哥們過生日，來了f個人，有n個披薩餅，這些披薩餅有著相同的厚度和各自的半徑。然後這些哥們想吃披薩，每個人又想吃的都是一樣的數量，而且不能大家都不想去吃用剩下的邊角料留下的披薩，所以就問每個人吃披薩餅的最大量。

思路大概就是二分答案了。但是有個小問題就是精度。首先是PI，要不用C++的反三角函數去計算，要不就去網上找常量去計算。接著是誤差限，一定要注意1e-5。最後一個就是提交問題了。我會回頭再說明選擇G++和C++的區別。

代碼示例

/****
	*@author    Shen
	*@title     poj 3122
	*/

#include 
#include 
#include 
using namespace std;

const double PI = acos(-1.0);
const double eps = 1e-5;

int n, f;
double r, v[10005];
double maxa = 0;

bool test(double x)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += int(v[i] / x);
    return sum >= (f + 1);
}

double Bsearch(double l, double r)
{
    while (r - l > eps)
    {
        double mid = (r + l) * 0.5;
        if (test(mid))
            l = mid;
        else r = mid;
    }
    return l;
}

void solve()
{
    scanf("%d%d", &n, &f);
    maxa = 0;
    for (int i = 0; i < n; i++)
    {
        scanf("%lf", &r);
        v[i] = r * r * PI;
        maxa = max(maxa, v[i]);
    }
    double ans = Bsearch(0.0, maxa);
    printf("%.4lf\n", ans);
}

int main()
{
    int t; scanf("%d", &t);
    while (t--)
        solve();
    return 0;
}