hdu 1969 &&nyoj 1193 Pie【二分法】

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6488    Accepted Submission(s): 2438

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

首先，这个题题意不太好懂.....英语不好，伤不起啊.......也可能是题意不明，说的就是啥吧，一个人有很多吃的东西，别管是啥喽，知道是吃的就行.....然后很多他的朋友来一起吃，当然还包括他自己，然后现在要求分成若干份，而且每个人都不愿意要切的零零碎碎的（废话！），也就是说零碎的部分就浪费了，现在要你来帮忙分饼，要求浪费的尽量少（给出了精度），而且每个人分的尽量多................

题解：

这样的题，只能用二分法来逼近最优值，也就是最终满足条件的每个人分得的最大面积（本题给出的是圆柱体的半径，厚度为1，可以直接当成面积来处理.....），会用二分法，这个题也基本没问题了.....

不过...这个题的精度真是坑....pi 的精确值取到小数点后十位都不对，要去到十五六位............汗.......

好在有一个比较常用的 pi 的定义方法，记下来，以后会经常用的哦~~

#include<stdio.h>
#include<math.h>
#define pi acos(-1)//记下来，就不会被坑了.....
double x[10005],max;
int n,f;
int sum(double mid)//统计分得的数量（整数哦.）
{
	int s=0;
	for(int i=0;i<n;++i)
	{
		s+=(int)(x[i]/mid);
	}
	return s>=f;
}
int main()
{
	int t,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&f);
		++f;max=0;
		for(i=0;i<n;++i)
		{
			scanf("%lf",&x[i]);
			x[i]=x[i]*x[i]*pi;
			if(x[i]>max)//先找最大的饼，因为按规则，每个人最多分得的不会超过最的饼的面积......
			{
				max=x[i];
			}
		}
		double l=0,r=max,mid;//开始二分...
		while(fabs(r-l)>1e-7)//精度太少会错误，太多会超时，无语了....
		{
			mid=(l+r)/2.0;
			if(sum(mid))//限定区间
			{
				l=mid;
			}
			else
			{
				r=mid;
			}
		}
		printf("%.4lf\n",l);// l 是满足题意而且最大的哦...
	}
	return 0;
}