Timer
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit:
32768/32768 K (Java/Others) Total Submission(s):
445 Accepted Submission(s): 90
Problem Description
Recently, some archaeologists discovered an ancient relic
on a small island in the Pacific Ocean. In the relic, they found an interesting
cone shaped container with transparent bottom. The container lay on the
horizontal ground with its central axis parallel to the ground. Just beside the
container, they found a manual telling them something about the container.
The container was a timer for a
special ceremony. Ancient people filled it all with water before the ceremony,
and when the ceremony began, they pulled out the plug in the small hole on the
tip of the cone to let the water out. There was a horizontal line called “sacred
line” carved on the bottom of the cone, and when the water level hit that line,
they pushed the plug back and ended the ceremony. But the archaeologists could
not found the sacred line on that cone. In order to sell the timer at a good
prize, the archaeologists wanted to recover that very important line.
By the manual they figured out
how much water flew out when the ceremony ended, but they don’t know what to do
next, so they come to you for help.
They measures the height of the
cone, and the diameter of the bottom, you should tell them the sacred line’s
height above the ground.
Input
The first line of the input contains an integer
T(1<=T<=20), indicating the number of test cases.
Each line after that
is a test case. It contains three real numbers, H, D(1<=H,D<=1000) and V,
indicating the height and bottom diameter of the timer, and the volume of water
that flew out during the ceremony. That volume is guaranteed to be less than
half volume of the container.
Output
For each test case, output one line containing the height
of the sacred line above the ground.
You should round off the answers to the
5th decimal place. (For example, rounding off 4.000005 equals to 4.00001 and
rounding off 4.000004 equals to 4.00000)
Sample Input
2 5.0 10.0 0.0 5.0
10.0 65.4498
Sample Output
10.00000 5.00000
Source
数学题： 看了结题报告做的
代码：
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<math.h>
4 const double ZERO =1e-8 ;
5 double H,D,V,R;
6 double calc(double r)
7 {
8 double h=R-r;
9 return H*R*R*acos(h/R)/3 -h*H*sqrt(R*R-h*h)*2/3+h*h*h*H/R*log((R+sqrt(R*R-h*h))/h)/3;
10 }
11 int main()
12 {
13 int cases =0;
14 double l,r,mid;
15 scanf("%d",&cases);
16 while(cases--)
17 {
18 scanf("%lf%lf%lf",&H,&D,&V);
19 R=D/2.0;
20 l=0; r=R;
21 while(r-l>=ZERO)
22 {
23 mid=(l+r)/2;
24 if(calc(mid)<V) l=mid;
25 else r=mid;
26 }
27 printf("%.5lf\n",2*R-(l+r)/2.0);
28 }
29 return 0;
30 }
HDUOJ-------2493Timer(数学 2008北京现场赛H题),布布扣,bubuko.com