HDU 2289 Cup【高精度，二分】

Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8942 Accepted Submission(s): 2744

Problem Description

The WHU ACM Team has a big cup, with which every member drinks water. Now, we know the volume of the water in the cup, can you tell us it height?

The radius of the cup's top and bottom circle is known, the cup's height is also known.

Input

The input consists of several test cases. The first line of input contains an integer T, indicating the num of test cases.
Each test case is on a single line, and it consists of four floating point numbers: r, R, H, V, representing the bottom radius, the top radius, the height and the volume of the hot water.

Technical Specification

1. T ≤ 20.
2. 1 ≤ r, R, H ≤ 100; 0 ≤ V ≤ 1000,000,000.
3. r ≤ R.
4. r, R, H, V are separated by ONE whitespace.
5. There is NO empty line between two neighboring cases.

Output

For each test case, output the height of hot water on a single line. Please round it to six fractional digits.

Sample Input

1

100 100 100 3141562

Sample Output

99.999024

Source

The 4th Baidu Cup final

Recommend

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2289

分析：二分，对高度进行二分，然后我们知道圆台的公式为V=pi/3*(r*r+u*u+r*u)*h;然后可二分解决！这题精度比较坑爹，所以稍微注意点就好了！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 inline int read()
 5 {
 6     int x=0,f=1;
 7     char ch=getchar();
 8     while(ch<'0'||ch>'9')
 9     {
10         if(ch=='-')
11             f=-1;
12         ch=getchar();
13     }
14     while(ch>='0'&&ch<='9')
15     {
16         x=x*10+ch-'0';
17         ch=getchar();
18     }
19     return x*f;
20 }
21 inline void write(int x)
22 {
23     if(x<0)
24     {
25         putchar('-');
26         x=-x;
27     }
28     if(x>9)
29     {
30         write(x/10);
31     }
32     putchar(x%10+'0');
33 }
34 const double pi=acos(-1.0);
35 const double eps=1e-9;
36 double r,R,h,H,V;
37 inline double check(double r,double R,double h,double H)
38 {
39     double u=h/H*(R-r)+r;
40     return pi*1.0*(r*r+u*u+u*r)*h/3;
41 }
42 int main()
43 {
44     int n;
45     n=read();
46     while(n--)
47     {
48         scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
49         double L1=0.0;
50         double R1=100.0;
51         double mid;
52         while(L1+eps<R1)
53         {
54             mid=(L1+R1)/2;
55             double vv=check(r,R,mid,H);
56             if(fabs(vv-V)<=eps)
57                 break;
58             else if(vv>V)
59                 R1=mid-eps;
60             else L1=mid+eps;
61         }
62         printf("%.6lf\n",mid);
63     }
64     return 0;
65 }