题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1969

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4513 Accepted Submission(s): 1819

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

4 3 3

1 24

10 5

1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327

3.1416

50.2655

题目大意：生日聚会上，来到的所有朋友都会领取一块饼，但是要求的是所有人都要拿到大小相同的饼！题目要求输出的是：每个人尽量拿到的最大的饼的体积是多大。这里的饼的形状是高为1的圆柱形。V=底面积*高

特别注意：1、题目里要给你自己留一块。

2、每个人得到的饼不可以是两块饼拼接在一起的。

3、这里的PI要用acos(-1.0)，具体原因还不知道为什么，可能精度要求比较高，因为这里wa几次。0.0

详见代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4
 5 using namespace std;
 6
 7 #define PI acos(-1.0)
 8
 9 int main ()
10 {
11     int t;
12     while (~scanf("%d",&t))
13     {
14         while (t--)
15         {
16             int n,f,r;
17             double V[10005],v=0,vmax;
18             scanf("%d%d",&n,&f);
19             f=f+1;
20             for (int i=0; i<n; i++)
21             {
22                 scanf("%d",&r);
23                 V[i]=r*r*PI;
24                 v+=V[i];
25                 //cout<<v<<endl;
26             }
27             vmax=v/f;
28             double left,right,mid;
29             left=0;
30             right=vmax;
31             int ans;
32             while ((right-left)>1e-6)
33             {
34                 int flag=0,k=0;
35                 ans=0;
36                 mid=(left+right)/2;
37                 for (int i=0;i<n;i++)
38                 {
39                    // ans+=(int)(V[i]/mid);
40                     //if (ans>=f)
41                         //flag=1;
42                     double vv=V[i];
43                     while(vv>=mid)
44                     {
45                         vv-=mid;
46                         k++;
47                         //cout<<vv<<" "<<k<<endl;
48                         if(k==f)
49                         {
50                             flag=1;
51                             break;
52                         }
53                     }
54                     if(flag==1) break;
55                 }
56                 if(flag==1)
57                     left=mid;
58                 else
59                     right=mid;
60             }
61             printf("%.4lf\n",mid);
62         }
63     }
64     return 0;
65 }

另外一种

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4
 5 using namespace std;
 6
 7 #define PI acos(-1.0)
 8
 9 int main ()
10 {
11     int t;
12     while (~scanf("%d",&t))
13     {
14         while (t--)
15         {
16             int n,f,r;
17             double V[10005],v=0,vmax;
18             scanf("%d%d",&n,&f);
19             f=f+1;
20             for (int i=0; i<n; i++)
21             {
22                 scanf("%d",&r);
23                 V[i]=r*r*PI;
24                 v+=V[i];
25                 //cout<<v<<endl;
26             }
27             vmax=v/f;
28             double left,right,mid;
29             left=0;
30             right=vmax;
31             int ans;
32             while ((right-left)>1e-6)
33             {
34                 int flag=0,k=0;
35                 ans=0;
36                 mid=(left+right)/2;
37                 for (int i=0;i<n;i++)
38                 {
39                     ans+=(int)(V[i]/mid);
40                     if (ans>=f)
41                         flag=1;
42                     if(flag==1) break;
43                 }
44                 if(flag==1)
45                     left=mid;
46                 else
47                     right=mid;
48             }
49             printf("%.4lf\n",mid);
50         }
51     }
52     return 0;
53 }

转载于:https://www.cnblogs.com/qq-star/p/4259387.html

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