Hdoj 1969 Pie、二分：【题解】-优快云博客

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20350 Accepted Submission(s): 7007

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

这是一个要注意精确度的二分。过不去的原因有几个坑点：

pi的取值：#define pi acos(-1.0)
2.没有设立一个精确值，而且要是double类型的，鬼知道我为什么一开始设置成了int型的，导致对照了好久没找出问题来。const double esp = 1e-5

我这道题的思路是，从已有的Pie中二分，面积范围为0~S[n-1]（最大的一块Pie的面积）。check函数的写法是当传入某一个mid值来，对每一块Pie进行划分，看能划分几块整的mid值，用count做计数器，`count += (int)(s[i] / mid)`计算数量，如果count >=fnum + 1，那么这个面积可行，继续增加面积。

AC代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define pi acos(-1.0)
using namespace std;
const double esp = 1e-5;
int fnum, n;
double ri[10005];
double s[10005];
// 此函数：传入mid(mid是面积)，判断所有pie可不可以分出const(count >= fnum + 1)个饼子 
bool check(double mid){
	int count = 0;
	for(int i = 0; i < n; i++){
		count += (int)(s[i] / mid);
	}
	return count >= fnum + 1;
}
int main(){
	int t;
	while(~scanf("%d", &t)){
	while(t--){
		scanf("%d %d", &n, &fnum);
		for(int i = 0; i < n; i++){
			scanf("%lf", &ri[i]);
			s[i] = pi * ri[i] * ri[i];
		}
		sort(s, s + n);
		// l和r代表的是最小和最大面积 
		double l = 0.0, r = s[n-1];
		while(l + esp < r){
			double 	mid = (l + r) / 2.0;		// 可以换成位运算，更节省时间
			if(check(mid)){
				l = mid;
			}	
			else{
				r = mid;
			}	
		}
	printf("%.4lf\n", l);
	}
	}
	return 0;
}