Cylinder Candy（积分+体积+表面积+旋转体）

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



经过这几天的思考，以及各种抓狂，我只能说数学老师就是牛，一语点了醒我！
好纠结，，，难道以后要背着高数跑。。。



这是我最初的想法，然后经过求证，体积并不要减去里面的圆柱；


过程已经写得很详尽了。。。

转载请注明出处：寻找&星空の孩子 
题目链接：http://acm.zju.edu.cn/onlinejudge/contestInfo.do?contestId=361

 1 #include<stdio.h>
 2 #include<math.h>
 3 #define PI acos(-1.0)
 4 int main()
 5 {
 6     int T;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         double r,h,d;
11         double v,s;
12         scanf("%lf%lf%lf",&r,&h,&d);
13         v=PI*PI*r*d*d+4.0/3*PI*d*d*d+2*PI*r*r*d+PI*h*((d+r)*(d+r));
14         s=2*(PI*PI*r*d+2*PI*d*d+PI*r*r+PI*h*(r+d));
15         printf("%.12lf %.12lf
",v,s);
16 
17     }
18     return 0;
19 }

View Code