acm课程练习2--1003

题目描述

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

题目大意

有F+1个人分N块蛋糕，每人只能分一块，且每人分到的大小必须相等

思路

随着分的蛋糕面积的增大，能分成的块数递减（注意，这不是一个线性的函数关系，因为蛋糕不能重新组合，所以会出现一块蛋糕切出相同面积的几块后，由于剩余面积不及前几块大，只能舍弃剩余面积的情况。这也是为什么不能简单地用总面积除以人数的原因）

由于有以上的逆序递减关系，因此可以用二分法来找出解。

这是一道二分法的水题。

AC代码

#include<iostream>
#include<cmath>
#include<iomanip>
#include<stdio.h>
#define max(a,b) (((a)>(b))?(a):(b))
using namespace std;
const double pi=acos(-1.0);
int main(){
    //freopen("date.in","r",stdin);
    //freopen("date.out","w",stdout);
    int N,T,renshu,tem1,sum;
    cin>>T;
    double maxMian,tem2,low,up;
    double mianji[10001];
    while(T--){
        up=0;
        cin>>N>>renshu;
        renshu++;
        for(int i=0;i<N;i++){
            cin>>tem1;
            mianji[i]=pi*tem1*tem1;
            up=max(mianji[i],up);
        }
        low=0;
        sum=0;
        while(up-low>1e-6){
            sum=0;
            tem2=(up+low)/2;
            for(int j=0;j<N;j++){
                sum+=((int)(mianji[j]/tem2));
            }
            if(sum>=renshu) low=tem2;
            else up=tem2;
        }
        cout<<fixed<<setprecision(4)<<tem2<<endl;
    }
}

来自为知笔记(Wiz)