CF#195(div2) B:Vasily the Bear and Fly

http://codeforces.com/contest/336/problem/B

 

One beautiful day Vasily the bear painted 2m circles of the same radiusR on a coordinate plane. Circles with numbers from1 to
m had centers at points(2R - R, 0),
(4R - R, 0),
..., (2Rm - R, 0), respectively. Circles with numbers from
m + 1 to 2m had centers at points(2R - R, 2R),
(4R - R, 2R),
...,(2Rm - R, 2R), respectively.

Naturally, the bear painted the circles for a simple experiment with a fly. The experiment continued form^{2 days. Each day of the experiment got its own unique number from0
to m2 - 1, inclusive.}

On the day number i the following things happened:

1. The fly arrived at the coordinate plane at the center of the circle with number (
   is the result of dividing number x by numbery, rounded down to an integer).
2. The fly went along the coordinate plane to the center of the circle number (
   is the remainder after dividing number x by numbery). The bear noticed that the fly went from the center of circlev to the center of circle
   u along the shortest path with all points lying on the border or inside at least one of the2m circles. After the fly reached the center of circleu,
   it flew away in an unknown direction.

Help Vasily, count the average distance the fly went along the coordinate plane during each of thesem^{2 days.}

Input

The first line contains two integers m, R (1 ≤ m ≤ 10^{5,1 ≤ R ≤ 10).}

Output

In a single line print a single real number — the answer to the problem. The answer will be considered correct if its absolute or relative error doesn't exceed10^- 6.

Sample test(s)Input

1 1

Output

2.0000000000

Input

2 2

Output

5.4142135624

Note

Figure to the second sample

題意：給出m和r，一共有兩排，每排m個圓圈，其序號確定，一共有m^2天，每天七點是 ，終點是

要求出m^2天裡所走的平均路程，注意每次路程必須最短

 

思路：根據給出的條件，我們可以知道，最終所走的路線就是以每個下面圈為起點去遍曆上面所有的點，當距離大於3的時候，斜線必然要走兩次，當時一直卡在這裡了。

 

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;double a[100005],s[100005],ans;int main(){    int i,m;    double x = sqrt(2.0),r;    while(~scanf("%d%lf",&m,&r))    {        ans = 0;        s[0] = 0;        a[1] = 2.0*r;//m為1時的走法        a[2] = 2.0*r+x*r;//m為2隻走一斜線        a[3] = 2.0*r+2.0*x*r;//從下到上，要跨越的縱向距離為3，必然要斜走兩次，這樣才是最短        for(i = 4; i<=m; i++)            a[i] = a[i-1]+2*r;//除了斜走之外，全部都是直走        for(i = 1; i<=m; i++)            s[i] = s[i-1]+a[i];//將所有點的走法加起來        for(i = 1; i<=m; i++)            ans+=(s[i]+s[m-i+1]-a[1])/m;//加上每次的走法，因為s的限制，必須加上對稱點再減去重複的部位            /*這裡以m為5,i為2為例，i = 2時，走法應該是：a2+a1+a2+a3+a4,即a1+2*a2+a3+a4            s[2] = a1+a2，但是這樣走不到3,4,5，加上對稱點s[4] = a1+a2+a3+a4,才能得到正確的走法            */        printf("%.10lf\n",ans/m);    }    return 0;}