2014
01-05

# Cup

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.

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.

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.

1
100 100 100 3141562

99.999024

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define PI acos(-1.0)
#define exp 1e-9
double solve(double r,double R,double h,double H)
{
double u = h/H*(R-r) + r;
return PI/3*(r*r+r*u+u*u)*h;
}
int main()
{
int t;
double r,R,H,V,mid,vv,f,l;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
f=0;
l=100;
while(l-f>exp)
{
mid=(l+f)/2;
vv=solve(r,R,mid,H);
if(fabs(vv-V)<=exp)
break;
else if(vv>V)
l=mid-exp;
else
f=mid+exp;
}
printf("%.6lf\n",mid);
}
return 0;
}

1. 样例输出和程序输出不吻合，修改一下样例输出吧。我用的是VC编译器，会提示我的i和j变量重复定义