Problem:
Score: 400 points

Problem Statement

Amidakuji is a traditional method of lottery in Japan.

To make an amidakuji, we first draw W parallel vertical lines, and then draw horizontal lines that connect them. The length of each vertical line is H+1 [cm], and the endpoints of the horizontal lines must be at 1, 2, 3, ..., or H [cm] from the top of a vertical line.

A valid amidakuji is an amidakuji that satisfies the following conditions:

No two horizontal lines share an endpoint.

The two endpoints of each horizontal lines must be at the same height.

A horizontal line must connect adjacent vertical lines.

Find the number of the valid amidakuji that satisfy the following condition, modulo 1\ 000\ 000\ 007: if we trace the path from the top of the leftmost vertical line to the bottom, always following horizontal lines when we encounter them, we reach the bottom of the K-th vertical line from the left.

For example, in the following amidakuji, we will reach the bottom of the fourth vertical line from the left.

Constraints

H is an integer between 1 and 100 (inclusive).

W is an integer between 1 and 8 (inclusive).

K is an integer between 1 and W (inclusive).

Input

Input is given from Standard Input in the following format:

H W K

Output

Print the number of the amidakuji that satisfy the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

1 3 2

Sample Output 1

1

Only the following one amidakuji satisfies the condition:

Sample Input 2

1 3 1

Sample Output 2

2

Only the following two amidakuji satisfy the condition:

Sample Input 3

2 3 3

Sample Output 3

1

Only the following one amidakuji satisfies the condition:

Sample Input 4

2 3 1

Sample Output 4

5

Only the following five amidakuji satisfy the condition:

Sample Input 5

7 1 1

Sample Output 5

1

As there is only one vertical line, we cannot draw any horizontal lines. Thus, there is only one amidakuji that satisfies the condition: the amidakuji with no horizontal lines.

Sample Input 6

15 8 5

Sample Output 6

437760187

Be sure to print the answer modulo 1\ 000\ 000\ 007.

Solution in Rust (1.15.1) (judge verdict: Accepted)

//  ____            _                  _                       _       _
// |  _ \ _   _ ___| |_    ___ _ __   | |_ ___ _ __ ___  _ __ | | __ _| |_ ___
// | |_) | | | / __| __|  / __| '_ \  | __/ _ \ '_ ` _ \| '_ \| |/ _` | __/ _ \
// |  _ <| |_| \__ \ |_  | (__| |_) | | ||  __/ | | | | | |_) | | (_| | ||  __/
// |_| \_\\__,_|___/\__|  \___| .__/___\__\___|_| |_| |_| .__/|_|\__,_|\__\___|
//                            |_| |_____|               |_|

//https://github.com/manta1130/Competitive_Programming_Template_Rust

#[macro_use]
mod input {

    use std;
    use std::io;

    const SPLIT_DELIMITER: char = ' ';

    #[macro_export]
    #[allow(unused_macros)]
    macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
}

    #[allow(dead_code)]
    pub fn input_line_str() -> String {
        let mut s = String::new();
        io::stdin().read_line(&mut s).unwrap();
        s.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn p<T>(t: T)
    where
        T: std::fmt::Display,
    {
        println!("{}", t);
    }

    #[allow(dead_code)]
    pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<Vec<T>> = Vec::new();

        for _ in 0..line {
            let vec_line = input_vector();
            v.push(vec_line);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector<T>() -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<T> = Vec::new();

        let s = input_line_str();
        let split_result = s.split(SPLIT_DELIMITER);
        for z in split_result {
            let buf = match z.parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector_row<T>(n: usize) -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v = Vec::with_capacity(n);
        for _ in 0..n {
            let buf = match input_line_str().parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    pub trait ToCharVec {
        fn to_charvec(&self) -> Vec<char>;
    }

    impl ToCharVec for String {
        fn to_charvec(&self) -> Vec<char> {
            self.to_string().chars().collect::<Vec<_>>()
        }
    }
}

use input::*;

const MOD: usize = 1_000_000_007;

fn gen_bits(l: usize) -> Vec<usize> {
    let mut v = vec![];
    for mut i in 0..2_usize.pow(l as u32) {
        let ibuf = i;
        let mut flag = true;
        while i != 0 {
            if i & 3 == 3 {
                flag = false;
                break;
            }
            i >>= 1;
        }
        if flag {
            v.push(ibuf);
        }
    }
    v
}

fn main() {
    let (h, w, k): (usize, usize, usize);
    input!(h, w, k);
    let mut dp = vec![vec![0; w + 2]; h + 2];
    let bitlist = gen_bits(w - 1);
    dp[1][0] = 1;
    for i in 1..h + 1 {
        for j in 0..w + 1 {
            for b in &bitlist {
                if (j == 0 || ((*b >> (j - 1) & 1) == 0)) && (*b >> j) & 1 == 0 {
                    dp[i + 1][j] += dp[i][j];
                    dp[i + 1][j] %= MOD;
                }
            }
        }
        for j in 0..w + 1 {
            for b in &bitlist {
                if j != 0 && (*b >> (j - 1)) & 1 == 1 {
                    if j != 0 {
                        dp[i + 1][j - 1] += dp[i][j];
                        dp[i + 1][j - 1] %= MOD;
                    }
                } else if (*b >> j) & 1 == 1 {
                    dp[i + 1][j + 1] += dp[i][j];
                    dp[i + 1][j + 1] %= MOD;
                }
            }
        }
    }
    println!("{}", dp[h + 1][k - 1]);
}