HDOJ1969 二分查找问题[总结]

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1343    Accepted Submission(s): 457

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

 

Source

NWERC2006

 

Recommend

wangye

 

 

 

 

code：

#include<iostream>
#include<cmath>
using namespace std;

#define eps 1e-6
#define pi acos(-1.0)

double data[10000];
int N,F;

int dep(double x)
{
    int count=0;
    int i;
    for(i=0;i<N;i++)
    {
        count+=int(data[i]/x);
    }
    if(count>=F)
        return 1;
    else
        return 0;
}

int main()
{
    int n;
    int i;    
    double r;
    double right,sum,left,mid;
    while(~scanf("%d",&n))
    {
        while(n--)
        {
            sum=0;
            scanf("%d%d",&N,&F);
            for(i=0;i<N;i++)
            {
                scanf("%lf",&r);
                data[i]=r*r*pi;
                sum+=data[i];
            }
            F++;
            left=0;
            right=sum/F;
            while((right-left)>eps)
            {
                mid=(right+left)/2.0;
                if(dep(mid))
                    left=mid;
                else
                    right=mid;
            }
            printf("%.4f\n",mid);
        }
    }
    return 0;
}

 

 

第一天，一上午做了三道简答的二分问题，以及三分，涉及到了一些函数的极值求导等问题。

 

二分问题的关键是控制好left,right,mid以及精度的问题，再加上判断的方法，基本上就差不多了。