[Wunder Fund Round 2016 (Div. 1 + Div. 2 combined)] - E Robot Arm

洛谷傳送門

Codeforces傳送門

題目大意

給你一個有$n$段的機械臂， 給出$m$個操作， 每次將一截機械臂伸長$l_i$或順時針旋轉$l$°。 在每次操作後你需要回答一開始最靠右點當前的位置。

解題分析

很好的一道線段樹題， 假設$[l-1,l]$是水平的， 維護$[l, r]$相對於$l$的位置和最右端機械臂的角度， 然後就可以用線段樹合併區間得到答案。

總複雜度$O(Nlog(N))$。

代碼如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <algorithm>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define MX 300500
#define db double
#define ls (now << 1)
#define rs (now << 1 | 1)
#define ll long long
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
int n, m;
template <class T> IN T sqr(T x) {return x * x;}
const db PI = std::acos(-1);
struct Node {db x, y, rig;} tree[MX << 2];
IN void pushup(R int now)
{
	db rlen = std::sqrt(sqr(tree[rs].x) + sqr(tree[rs].y));
	db rang = std::atan2(tree[rs].y, tree[rs].x);
	tree[now].x = tree[ls].x + rlen * std::cos(tree[ls].rig + rang);
	tree[now].y = tree[ls].y + rlen * std::sin(tree[ls].rig + rang);
	tree[now].rig = tree[ls].rig + tree[rs].rig;
}
void build(R int now, R int lef, R int rig)
{
	if (rig - lef == 1)
	{
		tree[now].x = 1, tree[now].y = 0, tree[now].rig = 0;
		return;
	}
	int mid = lef + rig >> 1;
	build(ls, lef, mid), build(rs, mid, rig);
	pushup(now);
}
IN void modify(R int now, R int lef, R int rig, R db dang, R db dlen, R int tar)
{
	if (lef + 1 == rig)
	{
		db len = std::sqrt(sqr(tree[now].x) + sqr(tree[now].y));
		db ang = std::atan2(tree[now].y, tree[now].x);
		len += dlen, ang += dang;
		tree[now].x = std::cos(ang) * len;
		tree[now].y = std::sin(ang) * len;
		tree[now].rig = ang;
		return;
	}
	int mid = lef + rig >> 1;
	if (tar <= mid) modify(ls, lef, mid, dang, dlen, tar);
	else modify(rs, mid, rig, dang, dlen, tar);
	pushup(now);
}
int main(void)
{
	int typ, pos, del;
	in(n), in(m); build(1, 0, n);
	W (m--)
	{
		in(typ), in(pos), in(del);
		if (typ & 1) modify(1, 0, n, 0, del, pos);
		else modify(1, 0, n, -1.0 * del / 180 * PI, 0, pos);
		printf("%.10lf %.10lf\n", tree[1].x, tree[1].y);
	}
}