poj1905

Expanding Rods
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 8376 Accepted: 2058

Description

When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion.
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.

Your task is to compute the distance by which the center of the rod is displaced.

Input

The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

Output

For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision.

Sample Input

1000 100 0.0001
            15000 10 0.00006
            10 0 0.001
            -1 -1 -1
            

Sample Output

61.329
            225.020
            0.000
            

Source

推一下公式

然后二分就可以了

可以二分的有很多

但是如果二分圆心角的话感觉特别简单


code

#include <cstdio>
#include 
<cstdlib>
#include 
<cstring>
#include 
<cmath>
#include 
<ctime>
#include 
<cassert>
#include 
<iostream>
#include 
<sstream>
#include 
<fstream>
#include 
<map>
#include 
<set>
#include 
<vector>
#include 
<queue>
#include 
<algorithm>
#include 
<iomanip>
using namespace std;
double l,ll,n,c;
int main()
{
    
double left,mid,right;
    
while(scanf("%lf%lf%lf",&l,&n,&c)!=EOF)
    {
        
if(l==-1&&n==-1&&c==-1break;
        
if(l==0||n==0||c==0)
        {
            printf(
"0.000\n");
            
continue;
        }
        ll
=l*(1+n*c);
        left
=0;
        right
=acos(-1.0);
        
//二分角度
        while(right-left>1e-12)
        {
            mid
=(left+right)/2;
            
if(mid*l>2*ll*sin(mid/2))
                right
=mid;
            
else left=mid;
        }
        printf(
"%.3lf\n",(1-cos(mid/2))*l/(2*sin(mid/2)));
    }
    
return 0;
}

posted on 2012-08-02 17:02 jh818012 阅读(186) 评论(0)  编辑 收藏 引用


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