hdu 2289 CUP

Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1089    Accepted Submission(s): 340

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

1100 100 100 3141562

 

Sample Output

99.999024

 

Source

The 4th Baidu Cup final

 一到水题，比赛时却让我给wa了8次。。数据真恶心。。。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int t;double  r, R, H, V;double pi=3.141592653;void Swap(double &a,double &b){    double t;    t=a;    a=b;    b=t;}int main(){    cin>>t;    while(t--)    {        cin>>r>>R>>H>>V;        double v=pi*(r*r+R*R+r*R)*H/3.0;        if(V>v)        {            printf("%.6lf\n",H);            continue;        }        if(r==R)        {            printf("%.6lf\n",V/(pi*r*r));            continue;        }        else if(r<R)        {                double h2=(r*H)/(R-r);                double h1=h2+H;                double v1=pi*r*r*h2/3.0;                double v2=pi*R*R*h1/3.0;                double v3=v1+(V);                double t1=pow(v3/v2,1.0/3);                double l1=(t1*h1);                printf("%.6lf\n",(l1-h2));        }        else        {                Swap(r,R);                double h2=(r*H)/(R-r);                double h1=h2+H;                double v1=pi*r*r*h2/3.0;                double v2=pi*R*R*h1/3.0;                double v3=v1+(v-V);                double t1=pow(v3/v2,1.0/3);                double l1=(t1*h1);                printf("%.6lf\n",h1-l1);        }    }    return 0;}