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
After these days of thinking, as well as a variety of crazy, I can only say that the math teacher is a cow, a language point woke me!
Good tangle,,, is not going to carry a high number of running ...
This is my initial idea, and then after verification, volume does not have to subtract the inside of the cylinder;
The process has been written in great detail ...
Reprint please specify the source:
Title Link: http://acm.zju.edu.cn/onlinejudge/contestInfo.do?contestId=361
1#include <stdio.h>2#include <math.h>3 #definePI ACOs (-1.0)4 intMain ()5 {6 intT;7scanf"%d",&T);8 while(t--)9 {Ten Doubler,h,d; One Doublev,s; Ascanf"%LF%LF%LF",&r,&h,&d); -V=pi*pi*r*d*d+4.0/3*pi*d*d*d+2*pi*r*r*d+pi*h* ((d+r) * (d+R)); -s=2* (pi*pi*r*d+2*pi*d*d+pi*r*r+pi*h* (r+d)); theprintf"%.12LF%.12lf\n", v,s); - - } - return 0; +}
View Code
Cylinder Candy (integral + volume + surface area + rotating body)