思路:由樣例可知,若exit和實際command在當前的父節點兩邊,則父節點一側的子樹節點必須全部遍歷一遍,纔會再次返回到父節點,從而轉向正確的方向。
故先求出exit的正確路徑,然後與command相比,
①若相同,則只需沿着command走,ans++
②若不相同,則ans += 父節點左側子樹的節點數
代碼如下:
#include <cstdio>
using namespace std;
#define N 51
#define LEFT 0
#define RIGHT 1
long long fac[N], sum[N];
int dir[N];
int main(){
int h;
long long n;
scanf("%d %I64d", &h, &n);
fac[0] = 1, sum[0] = 1;
for(int i = 1; i <= h; ++i)
fac[i] = fac[i - 1] * 2, sum[i] = sum[i - 1] + fac[i];
int level = h;
long long ans = 0, cur = 0;
int cmd = LEFT;
for(int i = h; i > 0; --i){
if(n > cur + fac[i - 1])
dir[i] = RIGHT, cur += fac[i - 1];
}
while(level){
if(cmd == dir[level]){
++ans, --level;
}
else
ans += sum[level - 1];
cmd = 1 - cmd;
}
printf("%I64d\n", ans);
return 0;
}