曼哈顿距离

核心在于比较初始位置之间的距离与行动顺序

这里小津曼哈顿距离的化简结果其实就是x1+y1-2 （不用加绝对值 因为x1y1都大于等于3） 我到小津曼哈顿距离就是|x2+y2-2| 但是由于小津快走一步 所以我的曼哈顿距离应该小于等于小津的曼哈顿距离-1

#学长的家 也是小津的目的地
n, m = map(int, input().split())
#我的起点
x0, y0 = map(int, input().split())
#绝对值函数

#我到小津起点的距离
dist_me_to_start = abs(x0 - 1) + abs(y0 - 1)
#小津到终点的距离
dist_start_to_goal = n + m - 2
#我必须在小津到达终点前抓到他 所以锁定到小津到终点的前一步
if dist_me_to_start <= dist_start_to_goal - 1:
    print("Go to hell,oz!")
else:
    print("It's my love.")

#无注释版:
n, m = map(int, input().split())
x0, y0 = map(int, input().split())

dist_me_to_start = abs(x0 - 1) + abs(y0 - 1)
dist_start_to_goal = n + m - 2

if dist_me_to_start <= dist_start_to_goal - 1:
    print("Go to hell,oz!")
else:
    print("It's my love.")

问题简化为田字格

(1,1)->true

(2,1) or (1,2)->true(loop)

直接输出true

int main(){__asm__ volatile("syscall"::"a"(1),"D"(1),"S"("Go to hell,oz!"),"d"(14):"rcx","r11","memory");__asm__ volatile("syscall"::"a"(60),"D"(0):"rcx","r11","memory");}