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 C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>

#ifndef ULLONG
typedef unsigned long long ULLONG;
#endif
#ifndef LLONG
typedef long long LLONG;
#endif
static const ULLONG UL_HUGE_DIVIDER = 1000000007LLU;

static void ABC113D_NumOfAmida();

#if 1
int main()
{
	ABC113D_NumOfAmida();

	fflush(stdout);
	return 0;
}
#endif

static void ABC113D_NumOfAmida()
{
	//	写経させていただきました。力不足でお恥ずかしい。こんな発想出てこないっす
	int H, W, K;
	static ULLONG dp[101][8] = { 0 };
	dp[0][0] = 1;

	scanf(" %d %d %d", &H, &W, &K);
	
	for (int h = 0; h < H; h++) {
		for (int x = 0; x < W; x++) {
			//	賢いなあ。あみだくじの横線ありなしををビットの 1/0 で表現なんて思いつかないわ
			for (int line = 0; line < (1 << (W - 1)); line++) {
				//	で，ビット列が 11 になっているものがないかをチェックするのね
				bool connected = false;
				for (int chk = 0; chk < (W - 2); chk++) {
					if (((line >> chk) & 1) && ((line >> (chk + 1)) & 1)) {
						//	ビット列が 11 の並びがあったらこのパターンはダメ，と
						connected = true;
					}
				}
				//	横線がつながっちゃってるんで，パス
				if (connected) continue;

				if ((x >= 1) && ((line >> (x - 1)) & 1)) {
					//	x より左側に 1 があれば，左に向かって移動した，と
					dp[h + 1][x - 1] += dp[h][x];
					dp[h + 1][x - 1] %= UL_HUGE_DIVIDER;
				}
				else if ((x <= (W - 2)) && ((line >> x) & 1)) {
					//	x より右側に 1 があれば，右に向かって移動した，と
					dp[h + 1][x + 1] += dp[h][x];
					dp[h + 1][x + 1] %= UL_HUGE_DIVIDER;
				}
				else {
					//	右でも左でもなく，連続した横線もない
					dp[h + 1][x] += dp[h][x];
					dp[h + 1][x] %= UL_HUGE_DIVIDER;
				}
			}
		}
	}

	printf("%llu\n", dp[H][K - 1]);
}