HDU 4798 Skycity

Skycity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 355    Accepted Submission(s): 92

Problem Description

The world's new tallest building is going to be built in Changsha, which will be called as "Skycity". The Skycity is going to be built as a circular truncated cone, radius of its bottom is marked as R, and radius of its top is marked as r, height of the building is marked as H, and there will be F floors with exact the same height in the whole building.
After construction of the building's skeleton, the construction team is going to construct the curtain wall using thousands of glass panes. The curtain wall is installed in each floor. When installing the curtain wall in a floor, first the construction team will measure the radius r' of the ceiling, then they will install the glass curtain wall as a regular prism which can exactly contain the ceiling circle. When constructing the glass curtain wall, all the glass pane has a minimum area requirement S, and amount of glass usage should be as little as possible.

As all the glass has exact the same thickness, so we can calculate the consumption of each glass pane as its area. Could you calculate the minimum total glass consumption?

 

Input

There will be multiple test cases. In each test case, there will be 5 integers R, r (10 ≤ r < R ≤ 10000), H (100 ≤ H ≤ 10000), F (10 ≤ F ≤ 1000) and S (1 ≤ S <× r × H ÷ F) in one line.

 

Output

For each test case, please output the minimum total glass consumption, an absolute error not more than 1e-3 is acceptable.

 

Sample Input

50 10 800 120 5300 50 2000 500 10

 

Sample Output

149968.3082196020.459

 

Source

2013 Asia Changsha Regional Contest

思路：每层天花板的半径能根据高度算出来，然后的话，可以证明正n边行的n越大总面积越小，所以就是找最大的n是的单面的面积不低于S就行了，二分会超时，其实能直接算出来的，详情请看代码。

代码：

#include <iostream>#include <vector>#include <algorithm>#include <string.h>#include <cstring>#include <stdio.h>#include <cmath>#include <math.h>#define rep(i,a,b) for(int i=(a);i<(b);++i)#define rrep(i,b,a) for(int i = (b); i >= (a); --i)#define clr(a,x) memset(a,(x),sizeof(a))#define LL unsigned long long#define eps 1e-10using namespace std;const double PI = 4.0 * atan(1.0);double R,r,H,F,S;int dcmp(double a,double b){    if (a < b - eps) return -1;    else if (a > b + eps) return 1;    else return 0;}double h,rx;double MinnumunArea(){    double wmin = S / h;    double arpha = atan2(wmin,rx * 2);    int m = max(3,(int)(PI / arpha + 1e-12));    arpha =  PI / m;    //double w = rx * tan(arpha) * 2;    //return w * h * m;    return rx * tan(arpha) * 2.0 * h * m;}int main(){    #ifdef ACM        freopen("in.txt", "r", stdin);    #endif // ACM    while (scanf("%lf%lf%lf%lf%lf",&R,&r,&H,&F,&S)==5) {        h = H / F;        double ans = 0;        double ss = 0;        for(LL i = 1; i <= F; ++i) {            rx = R - (R-r) / F * i;            ans += MinnumunArea();        }        printf("%.3lf\n",ans);    }}