hdu 2289 Cup

Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2357 Accepted Submission(s): 741

Problem DescriptionThe 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.

InputThe 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.

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

Sample Input

1100 100 100 3141562

Sample Output

99.999024

SourceThe 4th Baidu Cup final

Recommendlcy

#include<stdio.h>#include<math.h>int main(){    int t;    double r,R,H,v,bop,top,mid,t1,t2,vm;//如題意    double pi=acos(-1.0);//計算出pi    scanf("%d",&t);    while(t--)    {        scanf("%lf%lf%lf%lf",&r,&R,&H,&v);        bop=0;        top=H;        while(top-bop>1e-12)//對高度進行二分        {            mid=(top+bop)/2;            t1=((R-r)*(R-r)*mid*mid*mid*pi)/(3*H*H);            t2=r*(R-r)*mid*mid*pi/H;            vm=pi*r*r*mid+t1+t2;//推匯出圓台的體積公式            //V=pi*r^2*h+((R-r)^2*h^3*pi)/(3*H^2)+r*(R-r)*mid^2*pi/H;            if(vm>v)                top=mid;            else                bop=mid;        }        printf("%.6lf\n",bop);    }    return 0;}