Skycity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 269    Accepted Submission(s): 63

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 5
300 50 2000 500 10

Sample Output

149968.308

2196020.459

Source

2013 Asia Changsha Regional Contest

这条是2013年长沙的题目

二分的题目就是卡姿势，同浮点误差~

二分姿势稍有不当就WA

求第 i 层的圆的半径一层层的加又WA ， 要直接用公式  ri = add * i + r ;

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <cmath>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-8;
inline double getlen(int n,double r)
{
    return 2.0 * r * tan( PI / n );
}
int main()
{
    ios::sync_with_stdio(0);

    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL

    double R,r,H,F,S;
    while( cin >> R >> r >> H >> F >> S ){
        double h = H / F , ans = 0 , add = ( R - r ) / F , ri  ;

        for( int i = 0 ; i < F ; i++ )
        {
            ri = add * i  + r ;
            int ll = 3 ,rr = 100000 ;
            double area ;
            int num;
            while( ll <= rr )
            {
                int mm = ( ll + rr ) >> 1;
                double tmp = getlen( mm , ri ) * h;
                if( S + eps < tmp ){
                    area = tmp , num = mm;
                    ll = mm + 1;
                }
                else{
                    rr = mm - 1;
            　　}　　　　　　　 }
            ans += num * area;
        }

        printf("%.3lf\n",ans);
    }
    return 0;
}