ZOJ 3866 Cylinder Candy

Title Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5478

Surface:

Cylinder Candy Time limit: 2 Seconds Memory Limit: 65536 KB Special Judge

Edward the confectioner is making a new batch of chocolate covered candy. Each candy center are shaped as a cylinder with radius r mm and height h mm.

The candy center needs to is covered with a uniform coat of chocolate. The uniform coat of chocolate is d mm thick.

You is asked to calcualte the volume and the surface of the chocolate covered candy.

Input

There is multiple test cases. The first line of input contains an integer T (1≤t≤1000) indicating the number of test cases. For each test case:

There is three integers r , h in one line d . (1≤ r , h , d ≤100)

Output

For each case, print the volume and surface area of the "Candy in" line. The relative error should is less than 10-8.

Sample Input

21 1 11) 3 5

Sample Output

32.907950527415 51.1551353380771141.046818749128 532.235830206285

Solution:

Today's game, volume is a product, the surface area is not out of the dead, the evening asked a math department of the study of the younger brother, five minutes to take care of, really ashamed. The derivation process, after two days to send again.

First hair shape. A graph is a rotating body formed around the x-axis.

Code:

#include <iostream> #include <cmath> #include <iomanip> #define PI 4*atan (1.0) using namespace Std;int Main () {int t,r,h,d;    cin>>t;    while (t--)    {    cin>>r>>h>>d;    Cout<<fixed<<setprecision (<<2*) (2*D*D*D/3.0*PI+R*D*D*PI*PI/2) +pi* ((r+d) * (r+d)) * (h) +pi* (r*r ) *2*d<< "";    Cout<<fixed<<setprecision (<<2*) (pi*pi*d*r+2*pi*d*d) +2*pi*r*r+2*pi* (r+d) *h<<endl;    

ZOJ 3866 Cylinder Candy