Pie
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6752 Accepted Submission(s):
2564
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
Source
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一道二分的题，学了这个算法很久了，但一直没有独立的写出过代码，二分的题不容易想到用二分解决，所以需要仔细思考，不过这道题是很明显的二分，而且是利用精度解决问题。
题意：有n块蛋糕，m个人，自己也要蛋糕，所以是分给m+1个人，然后给出每块蛋糕的半径，每个人都要一块完整的蛋糕，则每个人最大能平均分到的蛋糕面积是多少。（蛋糕是平面圆）
附上代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 using namespace std;
7 int n,m;
8 double pi=acos(-1.0); //计算π的值，这样更精确
9 double a[10005];
10 int xx(double x)
11 {
12 int num=0;
13 for(int i=0; i<n; i++)
14 {
15 num+=int(a[i]/x); //计算有多少块蛋糕大于此时的平均面积
16 }
17 if(num>=m) //如果有这么多块面积分给所有人，则满足题意
18 return 1;
19 else
20 return 0;
21 }
22 int main()
23 {
24 int T,i,j,r;
25 double v;
26 scanf("%d",&T);
27 while(T--)
28 {
29 scanf("%d%d",&n,&m);
30 m++; //加上自己要的一块
31 for(i=0; i<n; i++)
32 {
33 scanf("%d",&r);
34 a[i]=r*r*pi; //求出每块蛋糕的面积
35 v+=a[i]; //记录所有蛋糕的总面积
36 }
37 double max,l,r,mid;
38 sort(a,a+n); //二分之前一定要排序
39 max=v/m; //每个人平均分到的最大蛋糕就是蛋糕总面积的平均值
40 l=0.0,r=max; //面积从0到平均值二分
41 while((r-l)>1e-6) //满足这个精度则跳出循环（不了解精度问题，模仿别人的写法）
42 {
43 mid=(r+l)/2; //二分取中
44 if(xx(mid)) //判断，若满足，则最大蛋糕面积可能更大，则取后半段继续搜
45 l=mid;
46 else //若不满足，则蛋糕取大了，则取前半段继续搜
47 r=mid;
48 }
49 printf("%.4lf\n",l); //保留四位小数输出
50 }
51 }