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package org.psStrategies;

import java.util.Scanner;

/*

* [문제]

* n,m크기의 맵에서 (a,b)위치에 보물이 있을때, 최단거리로 도착할경우의수와 최단거리 출력

*

* */

public class test5 {

static int n, m, a, b;

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

n = sc.nextInt();

m = sc.nextInt();

a = sc.nextInt();

b = sc.nextInt();

if ((a == 0 && b == 0) || n <= a || m <= b) {

System.out.println("fail");

} else {

solve();

}

}

static int[] dx = { 1, 0 };

static int[] dy = { 0, 1 };

private static void solve() {

boolean[][] visited = new boolean[a + 1][b + 1];

int[][] map = new int[a + 1][b + 1];

map[0][0] = 1;

for (int i = 0; i <= a; i++) {

for (int j = 0; j <= b; j++) {

for (int k = 0; k < 2; k++) {

int x = i + dx[k];

int y = j + dy[k];

if (x > a || y > b)

continue;

map[x][y] += map[i][j];

}

}

}

for (int i = 0; i <= a; i++) {

for (int j = 0; j <= b; j++) {

System.out.print(map[i][j] + " ");

}

System.out.println();

}

System.out.println(map[a][b] + "\n" + (a + b));

}

/*

장애물이 없고, 맵의 크기가 작으니까, 수학적 접근가능

1. 1로 초기화하고

dp[i][0] = 1

dp[0][1] = 1

2. dp[i][j] = dp[i-1][j] + dp[i][j-1] 이용.

0C0 0C1 0C2 0C3

1C0 1C1 1C2 1C3

2C0 2C1 2C2 2C3

3C0 3C1 3C2 3C3

* */

}