【OpenJudge 8463】Stupid cat & Doge

http://noi.openjudge.cn/ch0204/8463/
挺恶心的一道简单分治。
一开始准备非递归。
大if判断，后来发现代码量过长，决定大打表判断后继情况，后来发现序号不对称。
最后发现非递归分治非常不可做。
采用递归和坐标变换，降低了编程复杂度和思维复杂度。
坐标变换的思想十分的清晰啊！它是个好东西啊，以后要善用。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll in() {
	ll k = 0; char c = getchar();
	for (; c < '0' || c > '9'; c = getchar());
	for (; c >= '0' && c <= '9'; c = getchar())
		k = k * 10 + c - 48;
	return k;
}

void get(ll n, ll num, ll &x, ll &y) {
	if (n == 1) {
		if (num == 1)
			x = 1, y = 1;
		else if (num == 2)
			x = 1, y = 2;
		else if (num == 3)
			x = 2, y = 2;
		else
			x = 2, y = 1;
		return;
	}
	ll midb, mida, mid, right = 1LL << (n * 2), xx, yy;
	mid = right >> 1;
	midb = mid >> 1;
	mida = mid + midb;
	if (num <= midb) {
		get(n - 1, num, xx, yy);
		x = yy; y = xx;
	} else if (num <= mid) {
		get(n - 1, num - midb, xx, yy);
		x = xx; y = yy + (1LL << (n - 1));
	} else if (num <= mida) {
		get(n - 1, num - mid, xx, yy);
		x = xx + (1LL << (n - 1)); y = yy + (1LL << (n - 1));
	} else {
		get(n - 1, num - mida, xx, yy);
		x = (1LL << n) + 1 - yy; y = (1LL << (n - 1)) + 1 - xx;
	}
}

double sqr(double x) {return x * x;}

ll T, n, S, D, Sx, Sy, Dx, Dy;

int main() {
	T = in();
	while (T--) {
		n = in(); S = in(); D = in();
		get(n, S, Sx, Sy);// printf("%I64d %I64d ", Sx, Sy);
		get(n, D, Dx, Dy);// printf("%I64d %I64d\n", Dx, Dy);
		printf("%.0lf\n", sqrt(sqr(10.0 * (Sx - Dx)) + sqr(10.0 * (Sy - Dy))));
	}
	return 0;
}