SDUT-2878-Circle（概率DP/高斯消元）

Circle

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

You have been given a circle from 0 to n - 1. If you are currently at x, you will move to (x - 1) mod n or (x + 1) mod n with equal probability. Now we want to know the expected number of steps you need to reach x from 0.

输入

The first line contains one integer T — the number of test cases.

 

Each of the next T lines contains two integers n, x (0  ≤ x < n ≤ 1000) as we mention above.

输出

For each test case. Print a single float number — the expected number of steps you need to reach x from 0. The figure is accurate to 4 decimal places.

示例输入

3
3 2
5 4
10 5

示例输出

2.0000
4.0000
25.0000

提示

 

来源

2014年山东省第五届ACM大学生程序设计竞赛

示例程序

 E[p]=0.5*E[p-1]+0.5*E[p+1]+1

 即  -0.5*E[p-1]+E[p]-0.5*E[p+1]=1。

可以用高斯消元来做，

另外根据规律可以得到d[i]=(n-1)+(n-3)+（n-5）。。。

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SDUT-2878-Circle（概率DP/高斯消元）