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2013
12-26

hdu 1969 Pie[解题报告]C++

Pie

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

样例输出:

25.1327
3.1416
50.2655

/*
分析:
    晕了,π非要取acos(-1.0)才行么?都取3.1415927了,
还WA。。。

                                      2012-08-01 20:25
*/

#include"stdio.h"
#include"math.h"
#define PI acos(-1.0)


double MAX(double a,double b)
{
	return a>b?a:b;
}
int main()
{
	int T;
	int n,peo;
	int i;
	int temp;
	double low,up,mid;
	double S[10011];


	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&peo);
		peo++;


		up=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&temp);
			S[i]=PI*temp*temp;
			up=MAX(up,S[i]);
		}


		low=0;
		mid=(low+up)/2;
		while(up-low>=0.0000001)
		{
			temp=0;
			for(i=0;i<n;i++)	temp+=(int)(S[i]/mid);
			if(temp>=peo)	low=mid;
			else			up=mid;
			mid=(low+up)/2;
		}
		printf("%0.4lf\n",mid);
	}
	return 0;
}

解题转自:http://blog.csdn.net/ice_crazy/article/details/7819984


  1. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的

  2. 第一题是不是可以这样想,生了n孩子的家庭等价于n个家庭各生了一个1个孩子,这样最后男女的比例还是1:1