Broken robot CodeForces - 24D (概率DP)

You received as a gift a very clever robot walking on a rectangular board. Unfortunately, you understood that it is broken and behaves rather strangely (randomly). The board consists of N rows and M columns of cells. The robot is initially at some cell on the i-th row and the j-th column. Then at every step the robot could go to some another cell. The aim is to go to the bottommost (N-th) row. The robot can stay at it's current cell, move to the left, move to the right, or move to the cell below the current. If the robot is in the leftmost column it cannot move to the left, and if it is in the rightmost column it cannot move to the right. At every step all possible moves are equally probable. Return the expected number of step to reach the bottommost row.

Input

On the first line you will be given two space separated integers N and M (1 ≤ N, M ≤ 1000). On the second line you will be given another two space separated integers i and j (1 ≤ i ≤ N, 1 ≤ j ≤ M) — the number of the initial row and the number of the initial column. Note that, (1, 1) is the upper left corner of the board and (N, M) is the bottom right corner.

Output

Output the expected number of steps on a line of itself with at least 4 digits after the decimal point.

Examples

Input

10 10
10 4

Output

0.0000000000

Input

10 14
5 14

Output

18.0038068653


题意：一个N*M的矩阵，一个机器人站在x，y的位置，每次可以走四种：1.呆在原地 2.向下一步 3.向左一步 4.向右一步。问走到最后一行走的步数的期望是多少
题解：网上有许多高斯消元的方法，这道题也可以多次逼近写，直接推出转移方程。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e5+5;
const int  mod = 1e9+7;

double f[1005][1005];

int main()
{
    int N,M;
    scanf("%d %d",&N,&M);
    int x,y;
    scanf("%d %d",&x,&y);
    for(int i=N-1;i>=x;i--)
    {
        for(int k=1;k<=50;k++)
        {
            for(int j=1;j<=M;j++)
            {
                if(M == 1)
                    f[i][1] = (f[i][1] + f[i+1][1])/2 + 1;
                else if(j == 1)
                    f[i][1] = (f[i][1] + f[i][2] + f[i+1][1])/3 + 1;
                else if(j == M)
                    f[i][M] = (f[i][M] + f[i][M-1] + f[i+1][M])/3 + 1;
                else
                    f[i][j] = (f[i][j-1] + f[i][j] + f[i][j+1] + f[i+1][j])/4 + 1;
            }
        }
    }
    printf("%lf\n",f[x][y]);
}