80分求调，必关

#include<iostream>
using namespace std;
long long n,m,a[25][25],x,y; 
int main(){
	scanf("%lld%lld%lld%lld",&n,&m,&x,&y);
	n+=2;m+=2;x+=2;y+=2;
	a[x][y]=a[x-2][y-1]=a[x-2][y+1]=a[x-1][y+2]=a[x+1][y+2]=a[x+2][y-1]=a[x+2][y+1]=a[x-1][y-2]=a[x+1][y-2]=-1;
	for(int i=2;i<=m;i++){
		if(a[2][i]==-1)break;
		a[2][i]=1;
	}
	for(int i=2;i<=m;i++){
		if(a[i][2]==-1)break;
		a[i][2]=1;
	}
	for(int i=3;i<=n;i++){
		for(int j=3;j<=m;j++){
			if(a[i][j]==-1)continue;
			if(a[i-1][j]!=-1)a[i][j]+=a[i-1][j];
			if(a[i][j-1]!=-1)a[i][j]+=a[i][j-1];
		}
	}
	printf("%lld",a[n][m]);
	return 0;
}