Codeforces 507B Amr and Pins【思维】

Amr and Pins

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

Input

2 0 0 0 4

Output

1

Input

1 1 1 4 4

Output

3

Input

4 5 6 5 6

Output

0

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).

题目大意：

给你一个圆，其半径为R，我们想将圆的圆心从X，Y，移动到X',Y'.每次移动可以选择圆上一点，并且将圆旋转任意角度，问最少用多少步就能将圆心移动到X',Y'；

思路：

贪心的去走，每次我们都走2r长度的路径，如果不满2r长度的，也算走一步。

Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<iostream>using namespace std;#define ll __int64double r,x,y,x2,y2;int main(){    while(~scanf("%lf%lf%lf%lf%lf",&r,&x,&y,&x2,&y2))    {        double dis=sqrt((x2-x)*(x2-x)+(y2-y)*(y2-y));        double ans=dis/(2*r);        ll ans1=(ll)ans;        if(ans1!=ans)        {            ans1++;        }        printf("%I64d\n",ans1);    }}