ARC103B Robot Arms 题解

题面

题意

转成切比雪夫距离后容易发现有解的充要条件是所有 \(x+y\) 奇偶性一样。

观察到 \(m\le 40\)，大概是 \(\log(x+y)\) 的范围，于是可以考虑二进制拆分。

结论：\(1,2,4,\dots,2^k\) 可以拼出所有 \(|x|+|y|\le 2^{k+1}-1\) 且 \(x+y\equiv 1\pmod 2\) 的坐标。可以用递归证明。

那么直接构造答案，如果 \(x+y\) 为偶数就在集合中多加一个 \(1\)。

从大到小枚举集合中的每个数，每次选择 \(x\)，\(y\) 中绝对值较大的一个操作。

代码：

#include <bits/stdc++.h>
#define DC int T = gi <int> (); while (T--)
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair <int, int> PII;
typedef pair <LL, LL> PLL;

template <typename T>
inline T gi()
{
	T x = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int N = 1003, M = N << 1;

int n;
LL x[N], y[N];
bool fl1, fl2;
LL pw2[33];

template <typename T>
inline T mabs(T x) {return x < 0 ? -x : x;}

int main()
{
	//freopen(".in", "r", stdin); freopen(".out", "w", stdout);
	n = gi <int> ();
	for (int i = 1; i <= n; i+=1) x[i] = gi <LL> (), y[i] = gi <LL> (), fl1 |= ((x[i] + y[i]) % 2 == 0), fl2 |= ((x[i] + y[i]) & 1);
	if (fl1 && fl2) return puts("-1"), 0;
	for (int i = (pw2[0] = 1); i <= 31; i+=1) pw2[i] = 2ll * pw2[i - 1];
	int mx = 31;
	if (fl1)
	{
		++mx;
		for (int i = 32; i >= 1; i-=1) pw2[i] = pw2[i - 1];
		pw2[0] = 1;
		puts("33");
		for (int i = 32; ~i; i-=1) printf("%lld ", pw2[i]);
	}
	else
	{
		puts("32");
		for (int i = 31; ~i; i-=1) printf("%lld ", pw2[i]);
	}
	puts("");
	for (int i = 1; i <= n; i+=1)
	{
		for (int j = mx; ~j; j-=1)
			if (mabs(x[i]) > mabs(y[i]))
			{
				if (x[i] < 0) x[i] += pw2[j], cout << 'L';
				else x[i] -= pw2[j], cout << 'R';
			}
			else
			{
				if (y[i] < 0) y[i] += pw2[j], cout << 'D';
				else y[i] -= pw2[j], cout << 'U';
			}
		puts("");
	}
	return !!0;
}

启发：

• 构造题观察到数据范围为 \(\log\) 级 \(\rightarrow\) 二进制拆分。

标签：__,typedef,pw2,int,题解,Robot,ARC103B,gi,define
来源： https://www.cnblogs.com/xsl19/p/arc103b.html