POJ 1905 Expanding Rods

时间：2015-04-06 12:51:52 收藏：0 阅读：253

Expanding Rods

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12796		Accepted: 3299

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

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define maxn 25100000
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-7
double l,n,c;
double a,r;
bool ok(double a){
   if(l/sin(a)*a<(1+n*c)*l)return true;
   else return false;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%lf%lf%lf",&l,&n,&c)){
        if(l==-1||n==-1||c==-1)break;
           double ua=0;
           double ub=pi/2;
           double mid;
           /*
        while((ub-ua)>eps){
            mid=(ub+ua)/2;
            if(ok(mid))ua=mid;
            else ub=mid;
        }
        */
        
        for(int i=0;i<100;i++){
            mid=(ua+ub)/2;
            if(ok(mid))ua=mid;
            else ub=mid;
        }
        printf("%.3f\n",l/2/sin(mid)*(1-cos(mid)));
    }
}

要用while写法的话就得用下面哪种判断方法，不知道为什么，精度的问题吗？

转自http://blog.csdn.net/u013748887/article/details/24369543

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
const double esp=1e-5;//最小精度
int main()
{
    double L,n,c,s;
    double h;
    double r;
    double mid;
    while(scanf("%lf%lf%lf",&L,&n,&c)!=EOF)
    {
        if(L==-1&&n==-1&&c==-1)
        break;
        double low=0;
        double high=0.5*L;
        double mid;
        s=(1+n*c)*L;
        while(high-low>esp)//这里不能直接用hign>low,否则当low与high很接近时会陷入死循环
        {
            mid=(high+low)/2;
            r=(4*mid*mid+L*L)/(8*mid);
            if(2*r*asin(L/(2*r))<s)
            low=mid;
            else high=mid;
        }
        h=mid;
        printf("%.3lf\n",h);
    }
    return 0;
}

原文：http://blog.csdn.net/u013497977/article/details/44901005