HDOJ/HDU 4033 2011成都网络赛C题

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xieshimao

最新推荐文章于 2025-08-12 20:35:18 发布

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分类专栏：数学文章标签：网络 c distance integer output numbers

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本文介绍了一种计算平面中正多边形边长的方法，通过给定点到各顶点的距离来逆向求解边长。利用二分查找算法结合余弦定律计算内角总和，判断解的存在性。

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Regular Polygon

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1041 Accepted Submission(s): 295

Problem Description

In a 2_D plane, there is a point strictly in a regular polygon with N sides. If you are given the distances between it and N vertexes of the regular polygon, can you calculate the length of reguler polygon's side? The distance is defined as dist(A, B) = sqrt( (Ax-Bx)*(Ax-Bx) + (Ay-By)*(Ay-By) ). And the distances are given counterclockwise.

Input

First a integer T (T≤ 50), indicates the number of test cases. Every test case begins with a integer N (3 ≤ N ≤ 100), which is the number of regular polygon's sides. In the second line are N float numbers, indicate the distance between the point and N vertexes of the regular polygon. All the distances are between (0, 10000), not inclusive.

Output

For the ith case, output one line “Case k: ” at first. Then for every test case, if there is such a regular polygon exist, output the side's length rounded to three digits after the decimal point, otherwise output “impossible”.

Sample Input

  
   2
3
3.0 4.0 5.0
3
1.0 2.0 3.0

Sample Output

  
   Case 1: 6.766
Case 2: impossible

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

可以直接二分那个边长，然后通过计算内角和是否大于360来判断当前的边长是否合理

我的代码：

#include<stdio.h>
#include<math.h>
#include<algorithm>
#define PI acos(-1.0)
#define eps 1e-8

using namespace std;

double a[105];
int n;

double cal(double x)
{
	int i;
	double sita=0,tmp1,tmp2;
	for(i=1;i<=n;i++)
	{
		tmp1=a[i]*a[i]+a[i+1]*a[i+1]-x*x;
		tmp2=2*a[i]*a[i+1];
		sita=sita+acos(tmp1/tmp2);
	}
	return sita;
}

double solve(double left,double right)
{
	int k;
	double mid;
	for(k=1;k<=200;k++)
	{
		mid=(left+right)/2;
		if(cal(mid)>2*PI+eps)
		{
			right=mid;
		}
		else
		{
			left=mid;
		}
	}
	return left;
}

double max(double a,double b)
{
	if(a>b)
		return a;
	else
		return b;
}

double min(double a,double b)
{
	if(a>b)
		return b;
	else
		return a;
}

int main()
{
	int i,t,T;
	double ans,left,right;
	scanf("%d",&T);
	for(t=1;t<=T;t++)
	{
		left=0,right=99999;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%lf",&a[i]);
		a[n+1]=a[1];
		for(i=1;i<=n;i++)
		{
			right=min(right,a[i]+a[i+1]);
			left=max(left,fabs(a[i]-a[i+1]));
		}
		right=right-eps;
		if(left>right)
		{
			printf("Case %d: impossible\n",t);
			continue;
		}
		ans=solve(left,right);
		if(fabs(cal(ans)-2*PI)<eps)
			printf("Case %d: %.3lf\n",t,ans);
		else
			printf("Case %d: impossible\n",t);
	}
	return 0;
}