hdu 1969 pie 卡精度的二分

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12138    Accepted Submission(s): 4280

Problem 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

题目意思不难 很经典的二分题目 要多注意的就是得处理精度问题

我们这里用 acos(-1.0)来个圆周率赋值

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
const double pl=acos(-1.0);// 圆周率！
double s[20001];
bool check(double key,int len,int msize)
{
    int ret=0;
    for(int i=1;i<=len;i++)
    {
        int temp=floor(s[i]/key);
        ret+=temp;
    }
    if( ret>=msize+1 ) return 1;
    else return 0;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,f;
        cin>>n>>f;
        double mid;
        for(int i=1;i<=n;i++)
        {
            double x;
            cin>>x;
            s[i]=x*x*pl;
        }
        double l=0,r=pl*100000000;// 最大值要想清楚
        for(int i=1;i<=1000;i++)// 其实100次就可以控制好精度了
        {
            mid=(l+r)/2;
            if(check(mid,n,f)) l=mid;
            else r=mid;
        }
        printf("%.4lf\n",mid);
    }
    return 0;
}