Poj3122--pie (two-point accuracy problem)

Pie

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11178		Accepted: 3899		Special Judge

Description

My birthday is coming up and traditionally I ' m serving pie. Not just one pie, no, I had a number N of them, of various tastes and of various sizes. F of my friends is coming to my party and each of the them gets a piece of pie. This should is one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My Friends is 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 (and 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 is of the same size.

What's the largest possible piece size all of us can get? All the Pies is cylindrical in shape and they all has the same height 1, but the radii of the Pies can is different.

Input

One line with a positive integer:the number of test cases. Then to each test case:

• One line with a 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 is given as a floating point number with an absolute error of in most 10?3.

Sample Input

33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2

Sample Output

25.13273.141650.2655

Source

Northwestern Europe 2006

The main idea: give n pies, m+1 personal average of these pies, each piece, ask each person the most points.

Two cents can get the pie size. Double two points

while ((high-low) > EQs)
{
Mid = (Low+high)/2.0;
if (Solve (mid))
{
Low = mid;
last = mid;
}
Else
High = mid;
}

Note: 1 double subtraction (high-low) > EQs, EQS used for card precision, eqs too much precision reduced, eqs too small time to become high.

22 points does not add 0.0001, that is, low = mid High = Mid; This may be slower, but the accuracy will be high.

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath>using namespace std;    Define PI 3.1415926535898#define eqs 1e-5double s[11000]; int n, M;d ouble F (double x) {int k = (x+eqs) * 10000;    x = k * 1.0/10000; return x;}    int solve (double x) {int i, j, num = 0;        for (i = n-1; I >= 0 && (s[i]-x) > eqs; i--) {j = s[i]/x;        num + = j;    if (num >= m+1) return 1;    } if (num >= m+1) return 1; return 0;}    int main () {int T, I, K;    Double low, mid, high, last;            while (scanf ("%d", &t)! = EOF) {while (t--) {scanf ("%d%d", &n, &m);            for (i = 0; i < n; i++) scanf ("%lf", &s[i]);            Sort (s,s+n);            for (i = 0; i < n; i++) s[i] = S[i]*s[i]*pi;            Low = 0;            High = s[n-1]; while ((high-low) > EQs) {mid = (Low+high)/2.0;                    if (Solve (mid)) {low = Mid;                last = mid;            } else High = mid;        } printf ("%.4lf\n", last); }} return 0;}

Poj3122--pie (two-point accuracy problem)