poj-2403-cup
题目描述
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.
输入
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.
输出
For each test case, output the height of hot water on a single line. Please round it to six fractional digits.
样例输入
样例输出
提示
#include<stdio.h>
#include<math.h>
#define PI 3.14159265
int main()
{
int t;
double r,R,H,V,v0,h1;
double ans;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
if(r!=R)
{
h1=(R-r)/(H*r)-H; //补全的小圆锥A的高，用相似比算出来
v0=(PI*r*r*h1)/3; //小圆锥A的体积
ans=(H+h1)-(H+h1)*pow((((PI*r*r*(H+h1)/3)-V)/(PI*r*r*(H+h1)/3)),1.0/3); //热水倒入后的体积的高 可以用总的体积减去热血倒入后的体积形成的那个圆锥B与总的补起来后的大圆锥C用相似法比 得出小圆锥B的高 所求的高为大圆锥C高-小圆锥B的高
}
else
{
ans=V/(PI*(r*r));
}
if(ans>H) ans=H;
printf("%.6lf\n",ans);
}
}
return 0;
}
