【Jason's_ACM_解题报告】Pie

最新推荐文章于 2023-01-02 23:57:14 发布

SarahLeigh

最新推荐文章于 2023-01-02 23:57:14 发布

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分类专栏： Jason's 文章标签：问题求解策略二分法圆周率的精确表示小数的科学计数法表示

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本文是关于Jason的ACM解题报告，主题是为生日派对分配等体积的圆柱形派块。题目要求在给定数量的派和朋友人数下，找出能均分的最大派块体积，保持所有派块体积相同，且考虑了派的半径不同。输入和输出格式详细说明，并给出了样例输入和输出。作者提到了在计算圆周率时的精度问题，建议使用acos(-1.0)代替近似值。

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Pie

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 ≤ 10000: the number of pies and the number of friends.
• One line with N integers ri with 1 ≤ ri ≤ 10000: 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 oating 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

水题一枚，不过栽到了PI的赋值上，我傻呵呵的宏定义了一个3.1415926，殊不知，更精确的是acos(-1.0)也就是arccos(-1.0)，另外0.00001可以用1e-5来表示。

附代码如下：

#include<cmath>
#include<cstdio> 
#include<cstring>
#include<algorithm>

using namespace std;

#define MAXN (10000+5)
#define clr(x) (memset(x,0,sizeof(x)))
#define P (acos(-1.0))//(3.1415926)

int n,f;
double s[MAXN];

bool check(double size){
	int total=0;
	for(int i=0;i<n;i++){
		total+=floor(s[i]/size);
		if(total>=f)return true;
	}
	return false;
}

int main(){
	int total;
	scanf("%d",&total);
	while(total--){
		clr(s);
		double maxS=0;
		scanf("%d%d",&n,&f);f++;
		for(int i=0;i<n;i++){
			int r;
			scanf("%d",&r);
			s[i]=r*r*P;
			maxS=max(maxS,(double)s[i]);
		}
		
		double L=0,R=maxS;
		while((R-L)>0.00001){
			double MID=(L+R)/2+0.00001;
			if(check(MID))L=MID;
				else R=MID-0.00001;
		}
		printf("%.4lf\n",L);
	}
	return 0;
}