HDU 1969 Pie【二分】

【分析】

“虽然不是求什么最大的最小值（或者反过来）什么的……但还是可以用二分的，因为之前就做过一道小数型二分题（下面等会讲） 考虑二分面积，下界L=0，上界R=∑ni=1nπ∗ri2。对于一个中值x和一张半径r的饼来说，这张饼能够分出的最多分数显然是：⌊π∗r2x⌋，所以我们只需要求出∑ni=1⌊π∗ri2x⌋，判断其与f+1的关系即可 

特别要注意的一点：π的大小！，因为r≤104r≤104，所以r2≤108r2≤108，又因为题目要求精确到三位小数，所以ππ就要精确到10^-11，即π=3.14159265358

”

Pie

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

 

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【题意】：有n个饼，半径为ri（ri≤104ri≤104），分给m+1个人，要求每个人分到的面积是一样的，而且每个人分到的饼是连续的一块（即不能是几个小块凑起来的），求能够分到的最大面积，精确到三位小数

 

【代码】：

#include<bits/stdc++.h>
using namespace std;
#define Lint long long int
#define  pi acos(-1)
const double eps=1e-6;
const int MAXN=10010;
double r[MAXN],s[MAXN];
double n,L,R;
int T,m;
bool check(double mid)
{
    int ret=0;
    for(int i=1;i<=n;i++)
        ret+=(int)(s[i]/mid);
    return ret>=m; //一旦出现Pie的数目大于等于到场人数的话就true 
}
int main()
{
    scanf("%d",&T);
    while( T-- )
    {
        L=R=0;
        scanf("%lf%d",&n,&m),m++;
        //m的原来数值只是朋友的个数，而主人公也是一员，所以加1，加上主人公   
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&r[i]);
            s[i]=pi*r[i]*r[i];   //这里是直接把所输入的半径变量变成面积，方便下面计算   
            R+=s[i];            // 上界 也可以写成R=max(R,r[i]);这里是吧右边界给设成pie里面的最大的那一个   
        }
        while( R-L>eps )
        {
            double mid=(L+R)/2;
            if( check( mid ) ) 
            //如果返回值为真，那么代表我们可以用此时mid的大小来替换左边界，看是否存在比mid还大但仍然满足所有人都能吃到一样大小的pie，因为题目上说的是尽可能大     
                L=mid;
            else   
            //一旦发现不符合的话，此时mid的大小就得替换掉右边界，继续在比mid小的大小里面找符合题目条件的大小   
                R=mid;
        }
        //最后输出l,因为只有满足条件的时候L的值才变，所以对于L来说，它是永远符合题目条件的那个值   
        printf("%.4lf\n",L);
    }
    return 0;
}