Zhejiang University 15th Annual school Race G-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 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

The game, to the last minute to calculate the volume, and still rely on teammate assists, but it was not done with calculus, after all, liberal arts, calculus or a little weak, after the great God guidance, finally pushed out.

Here is the derivation process, the curve surface area of the formula can only be memorized, I do not know how to prove.

The formula is deduced, the code is good to run ~ ~ ~

#include <cstdio> #include <cstring> #include <iostream> #include <cmath>using namespace std;# Define ll long longconst double PI = ACOs ( -1); int main () {    double R, H, D;    int T;    scanf ("%d", &t);    while (t--)    {        scanf ("%lf%lf%lf", &r, &h, &d);        Double V = 2*pi* (2.0/3*d*d*d + r*r*d + 0.5*pi*d*d*r) +pi* (r+d) * (r+d) *h;        Double S = 2*pi* (2*d*d+pi*d*r) + 2*pi*r*r + 2*pi* (r+d) *h;        printf ("%.12f%.12f\n", V, S);    }    return 0;}

To learn maths well ...

Zhejiang University 15th Annual school Race G-cylinder Candy