AcWing98 分形之城 (递归)
题目链接:https://www.acwing.com/problem/content/100/
为方便运算,城市编号从\(0\)开始,
按城市分级递归求解\(D\)编号城市的坐标
注意变换后坐标的计算(坐标从\((0,0)\)开始)
四舍五入用\(%0lf\)即可
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
const int maxn = 40;
int T;
int N;
ll A,B;
double dis(P a, P b){ return (double)sqrt(1.0 * (a.first - b.first) * (a.first - b.first) + 1.0 * (a.second - b.second) * (a.second - b.second)) * 10; }
P dfs(int n,ll pos){
if(n == 0){ return make_pair(0, 0); }
ll len = 1ll << (n - 1), num = 1ll << (2 * n - 2);
P tp = dfs(n - 1, pos % num);
ll x = tp.first, y = tp.second;
int d = pos / num;
if(d == 0) return make_pair(y, x);
if(d == 1) return make_pair(x, y + len);
if(d == 2) return make_pair(x + len, y + len);
if(d == 3) return make_pair(2 * len - y - 1, len - x - 1);
}
ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }
int main(){
T = read();
while(T--){
N = read(), A = read(), B = read(); --A, --B;
++N;
ll num = 1ll << (2 * N - 2);
P a = dfs(N - 1, A % num);
P b = dfs(N - 1, B % num);
// printf("%lld %lld %lld %lld\n",a.first, a.second, b.first, b.second);
printf("%.0lf\n",dis(a, b));
}
return 0;
}