poj1905
Expanding Rods
Description ![]() 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
但是如果二分圓心角的話感覺特別簡單
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==-1) break;
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;
}
#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==-1) break;
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;
}