HDU 1969 高精度



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

思路：二分法，但此题对精度要求很高

#include<stdio.h>#include<cstdlib>#include<math.h>//#define pi acos(-1.0）                  //3.1415926过不了，此题对精度要求很高//#define exp                         1e-7//时间复杂度的大小也与精度的大小有关int N,F;using namespace std;int str[10005];int judge(double mid){   long long  sum=0;    for(int i=1;i<=N;i++)            sum+=(int)(1.0*str[i]*str[i]*pi/mid);     return sum>=(F+1)? 1:0;//关键是在等于时候的判断}int main(){    //freopen("e://in.txt","r",stdin);    int V;    scanf("%d",&V);    while(V--)    {        double   sum=0.0;        scanf("%d%d",&N,&F);        for(int i=1;i<=N;i++)          {              scanf("%d",&str[i]);              sum+=1.0*str[i]*str[i]*pi;          }  sum/=1.0*(F+1);          double left=0.0,right=sum,mid;          while(right-left>=exp)          {             mid=(left+right)/2.0;            if(judge(mid))                left=mid;             else                right=mid;          }          printf("%.4lf\n",left);    }}
总结：1.pi =acos（-1.0），3.1415926不够精确

     2.精度也影响着时间复杂度