POJ - 3122 Pie

Pie

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status

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

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

Sample Output

25.13273.141650.2655

二分

#include<iostream>#include<iomanip>#include<cstdio>using namespace std;const double Pi = 3.14159265359;   //注意精度const double esp = 1e-6;int main(){int casen;cin >> casen;while (casen--){int n, f;double v[10005];          //每个pie的sizescanf_s("%d%d", &n, &f);f++;double maxsize = 0.0;for (int i = 1; i <= n; i++){scanf_s("%lf", &v[i]);v[i] = v[i] * v[i];if (v[i] > maxsize)        //求二分法的上界maxsize = v[i];}double high = maxsize;        //上界，每人都得到整个pie，而且那个pie为所有pie中最大的double low = 0.0;             //下界，每人都分不到piedouble mid;while (high - low > esp)       //实数double的二分结束条件不同于整数int的二分结束条件{mid = (high + low) / 2;   //对当前上下界折中，计算"如果按照mid的尺寸分pie，能分给多少人"int count_f = 0;           //根据mid尺寸能分给的人数for (int i = 1; i <= n; i++)//枚举每个piecount_f += (int)(v[i] / mid);  //第i个pie按照mid的尺寸去切，最多能分的人数（取整）if (count_f < f)  //当用mid尺寸分，可以分的人数小于额定人数high = mid; //说明mid偏大，上界优化else low = mid;  //否则mid偏小，下界优化（注意'='一定要放在下界优化，否则精度会出错）}cout << fixed << setprecision(4) << mid*Pi << endl; //之前的计算都只是利用半径平方去计算，最后的结果要记得乘pi}}