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最新推荐文章于 2020-04-21 16:28:02 发布

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探讨如何在保证公平的前提下，将不同大小的圆形派均匀分配给一群朋友，以求得每个人能获得的最大派体积。通过编程实现，利用二分查找算法确定最优解。

题目
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

题意：有一个人要给好朋友们分pie吃，每个人分到的pie大小要一样，而且每个人只能有一块，求每个人分到的最大体积。

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const double pi=acos(-1.0);
double s[10010];
int n,f;
int check(double mid)  //判断当前体积是否能保证每个人都能吃到pie
{
	int sum=0;
	for(int i=0;i<n;i++)
	{
		sum+=int(s[i]/mid);  //每块pie按mid分能分成几部分
		if(sum>=f) 
		    return 1;
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(s,0,sizeof(s));  //有多组输入，所以要清空数组
		scanf("%d%d",&n,&f);
		for(int i=0;i<n;i++)
		{
			scanf("%lf",&s[i]);
			s[i]=s[i]*s[i]*pi;  //直接求出每个pie的体积
		}
		f+=1;  //f为朋友的人数，要加上自己
		sort(s,s+n);  //给pie的大小排序
		double l=0,r=s[n-1];  //确定左右边界
		while(r-l>0.00001)  //这是题目中的范围限制
		{
			double mid=(l+r)/2;
			if(check(mid)) l=mid;  //当前体积够分，再次增大
			else r=mid;  //不够，减小
		}
		printf("%.4lf\n",l);
	}
	return 0;
}