B. Amr and Pins

最新推荐文章于 2018-09-15 17:26:14 发布

原创最新推荐文章于 2018-09-15 17:26:14 发布 · 642 阅读

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====ACM算法学习==== 同时被 3 个专栏收录

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贪心

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===== codeforces =====

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本文探讨了一个有趣的几何问题：如何通过围绕圆的边界上的点旋转来最小化圆心从初始位置到目标位置所需的步骤数量。文章提供了一种计算解决方案，并附带了实现该算法的C++代码。

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Amr loves Geometry. One day he came up with a very interesting problem.

Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').

In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.

Help Amr to achieve his goal in minimum number of steps.

Input

Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 10⁵, - 10⁵ ≤ x, y, x', y' ≤ 10⁵), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.

Output

Output a single integer — minimum number of steps required to move the center of the circle to the destination point.

Sample test(s)

Input

2 0 0 0 4

Output

Input

1 1 1 4 4

Output

Input

4 5 6 5 6

Output

Note

In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).

$\text{[math]}$

题意：

给你一个圆，一个初始圆心，一个目标圆心。

每次可以进行的变换为：围绕初始圆边上某一点，进行旋转，问最少经过多少次旋转，可以将圆旋转至目标位置。

分析：

也是贪心。。。计算初始圆心和目标圆心的距离，与圆直径相除，向上取余即为结果

AC：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <math.h>
using namespace std;
#define N 10010
#define LL __int64

int main(){
	//freopen("in.in","r",stdin);
	//freopen("out.out","w",stdout);
	double r,x,y,x1,y1;
	scanf("%lf %lf %lf %lf %lf",&r,&x,&y,&x1,&y1);
	double  t = sqrt((x1-x)*(x1-x)+(y1-y)*(y1-y))/(2*r);
	if(t-(int)t>1e-20) printf("%d\n",(int)t+1);
	else printf("%d\n",(int)t);

	//写的简单点，可以使用ceil(t)进一法进行取整。

	return 0;
}