emm,此题打表最合适
by Galaxy2aa @ 2023-10-01 12:59:57
@[Galaxy722](/user/1030479) 我打表了呀
by Michelle01 @ 2023-10-01 16:10:15
```
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int fx[] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
const int fy[] = {0, 1, 2, 2, 1, -1, -2, -2, -1};
//马可以走到的位置
int bx, by, mx, my;
ll f[40][40];
bool s[40][40]; //判断这个点有没有马拦住
int main(){
scanf("%d%d%d%d", &bx, &by, &mx, &my);
bx += 2; by += 2; mx += 2; my += 2;
//坐标+2以防越界
f[2][1] = 1;//初始化
s[mx][my] = 1;//标记马的位置
for(int i = 1; i <= 8; i++) s[mx + fx[i]][my + fy[i]] = 1;
for(int i = 2; i <= bx; i++){
for(int j = 2; j <= by; j++){
if(s[i][j]) continue; // 如果被马拦住就直接跳过
f[i][j] = f[i - 1][j] + f[i][j - 1];
//状态转移方程
}
}
printf("%lld\n", f[bx][by]);
return 0;
} ```
看看题解吧！
by Galaxy2aa @ 2023-10-02 11:14:58