#include<bits/stdc++.h>
using namespace std;
bool ma(int x,int y,int mx,int my)
{
    if((abs(x-mx)==1 && abs(y-my)==2) || (abs(y-my)==1 && abs(x-mx)==2) || (y==my && x==mx))
        return 1;
    return 0;
  
}
long long f[25][25];
int main()
{
    int x,y,mx,my;cin >> x >> y >> mx >> my;
    for(int i=0;i<=x;i++)
    {
        for(int j=0;j<=y;j++)
        {
        if(ma(i,j,mx,my))
            f[i][j] = 0;//马可到达
        else if(j==0 && i==0)
            f[i][j] = 1;//坐标原点
        else if(j==0)
            f[i][j] = f[i-1][j];//左边缘
        else if(i==0)
            f[i][j] = f[i][j-1];//上边缘
        else
            f[i][j] = f[i-1][j]+f[i][j-1];//中间
        }
    }
    cout << f[x][y];
    return 0;
}

一定用long long否则60分

#include <bits/stdc++.h>
long long bx,by,mx,my,a[25][25];
int horse(int x,int y){
    if(abs(x-mx)==2&&abs(y-my)==1||
       abs(x-mx)==1&&abs(y-my)==2)return 1;
    if(x==mx&&y==my)return 1;
    return 0;}
int main(){
    std::cin>>bx>>by>>mx>>my;
    for(int i=0;i<=bx;i++){
        for(int j=0;j<=by;j++){
            if(horse(i,j))a[i][j]=0;
            else if(i==0&&j==0)a[i][j]=1;
            else if(i==0)a[i][j]=a[i][j-1];
            else if(j==0)a[i][j]=a[i-1][j];
            else a[i][j]=a[i][j-1]+a[i-1][j];}}
    std::cout<<a[bx][by]; return 0;}

#include<iostream>
int bx,by,mx,my;
long long f[21][21];
bool check(int x,int y)
{
    if(abs(x-mx)==1&&abs(y-my)==2||abs(x-mx)==2&&abs(y-my)==1)return true;
    else if(x==mx&&y==my)return true;
    else return false;
}
int main()
{
    std::cin>>bx>>by>>mx>>my;
    for(int i=0;i<=bx;i+=1)for(int j=0;j<=by;j+=1)if(check(i,j))f[i][j]=0;else if(i==0&&j==0)f[i][j]=1;else if(i==0)f[i][j]=f[i][j-1];else if(j==0)f[i][j]=f[i-1][j];else f[i][j]=f[i-1][j]+f[i][j-1];
    std::cout<<f[bx][by];
}