解题思路：
用递归遍历每一种情况，并且根据马的位置去掉不能走过的点
注意事项:
参考代码:
#include <stdio.h>
int computer(int x1, int y1, int x, int y, int a, int b);
int main(void)
{
int a, b, x, y;
int count;
scanf("%d%d%d%d", &x, &y, &a, &b);
count = computer(0, 0, x, y, a, b);
printf("%d", count);
return 0;
}
int computer(int x1, int y1, int x, int y, int a, int b)
{
if (x1 == x && y1 == y)
{
return 1;
}
else if (x1 != x && y1 == y)
{
if (x1 == a + 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a - 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a + 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a - 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a && y1 == b)
{
return 0;
}
else
return computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == x && y1 != y)
{
if (x1 == a + 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b);
}
else if (x1 == a - 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b);
}
else if (x1 == a + 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b);
}
else if (x1 == a - 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b);
}
else if (x1 == a && y1 == b)
{
return 0;
}
else
return computer(x1, y1 + 1, x, y, a, b);
}
else
{
if (x1 == a + 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b) + computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a - 2)
{
if (y1 == b + 1 || y1 == b - 1)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b) + computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a + 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b) + computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a - 1)
{
if (y1 == b + 2 || y1 == b - 2)
return 0;
else
return computer(x1, y1 + 1, x, y, a, b) + computer(x1 + 1, y1, x, y, a, b);
}
else if (x1 == a && y1 == b)
{
return 0;
}
else
return computer(x1, y1 + 1, x, y, a, b) + computer(x1 + 1, y1, x, y, a, b);
}
}
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