ZOJ 3866 Cylinder Candy

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 is shaped as a cylinder with radius r mm and height h mm.

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

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

Input

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

Output

For each case, print the volume and surface area of the candy in one line. The relative error should be less than 10^-8.

Sample Input

2
1 1 1
1 3 5

Sample Output

32.907950527415 51.155135338077
1141.046818749128 532.235830206285

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3866

几何题，加了外层之后可以看作是上下两个旋转体加上中间的圆柱体(半径为r+d)，用积分算一下得到公式。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <string>
 5 #include <iomanip>
 6 #include <cmath>
 7 using namespace std;
 8 #define pai acos(-1.0)
 9 int T;
10 double r, h, d;
11 int main(){
12     scanf("%d", &T);
13     while(T--){
14         scanf("%lf%lf%lf", &r, &h, &d);
15         double v, area;
16         v = (pai*r*d*d*asin(1.0) + pai*(2.0/3.0*d*d*d + r*r*d))*2 + (r+d)*(r+d)*pai*h;
17         area = 2*(2*pai*d*d + 2*pai*r*d*asin(1.0)) + 2*pai*r*r + 2*pai*(r+d)*h;
18         printf("%.12lf %.12lf
", v, area);
19     }
20     
21     return 0;
22 }