Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) E. Robot Arm 线段树

E. Robot Arm

题目连接：

http://www.codeforces.com/contest/618/problem/E

Description

Roger is a robot. He has an arm that is a series of n segments connected to each other. The endpoints of the i-th segment are initially located at points (i - 1, 0) and (i, 0). The endpoint at (i - 1, 0) is colored red and the endpoint at (i, 0) is colored blue for all segments. Thus, the blue endpoint of the i-th segment is touching the red endpoint of the (i + 1)-th segment for all valid i.

Roger can move his arm in two different ways:

He can choose some segment and some value. This is denoted as choosing the segment number i and picking some positive l. This change happens as follows: the red endpoint of segment number i and segments from 1 to i - 1 are all fixed in place. Imagine a ray from the red endpoint to the blue endpoint. The blue endpoint and segments i + 1 through n are translated l units in the direction of this ray.

In this picture, the red point labeled A and segments before A stay in place, while the blue point labeled B and segments after B gets translated.

He can choose a segment and rotate it. This is denoted as choosing the segment number i, and an angle a. The red endpoint of the i-th segment will stay fixed in place. The blue endpoint of that segment and segments i + 1 to n will rotate clockwise by an angle of a degrees around the red endpoint.

In this picture, the red point labeled A and segments before A stay in place, while the blue point labeled B and segments after B get rotated around point A.

Roger will move his arm m times. These transformations are a bit complicated, and Roger easily loses track of where the blue endpoint of the last segment is. Help him compute the coordinates of the blue endpoint of the last segment after applying each operation. Note that these operations are cumulative, and Roger's arm may intersect itself arbitrarily during the moves.

Input

The first line of the input will contain two integers n and m (1 ≤ n, m ≤ 300 000) — the number of segments and the number of operations to perform.

Each of the next m lines contains three integers xi, yi and zi describing a move. If xi = 1, this line describes a move of type 1, where yi denotes the segment number and zi denotes the increase in the length. If xi = 2, this describes a move of type 2, where yi denotes the segment number, and zi denotes the angle in degrees. (1 ≤ xi ≤ 2, 1 ≤ yi ≤ n, 1 ≤ zi ≤ 359)

Output

Print m lines. The i-th line should contain two real values, denoting the coordinates of the blue endpoint of the last segment after applying operations 1, ..., i. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker program will consider your answer correct if for all coordinates.

Sample Input

5 4

1 1 3

2 3 90

2 5 48

1 4 1

Sample Output

8.0000000000 0.0000000000

5.0000000000 -3.0000000000

4.2568551745 -2.6691306064

4.2568551745 -3.6691306064

Hint

题意

有一个机械臂，这个机械臂上面有n个点，i个点和i+1个点通过一条线段相连接

有两个操作:

1 x y，将第x个线段伸长y米

2 x y,将x个线段绕着上一个线段顺时针旋转y°

然后每次询问之后，问你最后那个点的位置在哪儿

题解：

线段树

我们在线段树的每个叶子节点，维护当前这个线段较上一个线段的角度，和当前线段端点的坐标

修改很简单，直接暴力修改这个叶子节点就好了

push_up的时候，就是把两个线段融合成一个线段

这个大概是个初中数学，大概画一画就出来了吧（误

总之就是这样

还有一种做法是矩阵，旋转和延长，其实都是矩阵变换

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+7;
const double pi = acos(-1.0);
typedef double SgTreeDataType;
struct treenode
{
  int L , R  ;
  SgTreeDataType x,y,ang;
};

treenode tree[maxn*4];

inline void push_down(int o)
{

}

inline void push_up(int o)
{
    tree[o].x = tree[o*2].x+tree[o*2+1].x*cos(tree[o*2].ang)-tree[o*2+1].y*sin(tree[o*2].ang);
	tree[o].y = tree[o*2].y+tree[o*2+1].x*sin(tree[o*2].ang)+tree[o*2+1].y*cos(tree[o*2].ang);
	tree[o].ang = tree[o*2].ang+tree[o*2+1].ang;
}

inline void build_tree(int L , int R , int o)
{
	tree[o].L = L , tree[o].R = R;
	if (L == R)
    {
        tree[o].x = 1;
        tree[o].y = 0;
        tree[o].ang = 0;
        return;
    }
	if (R > L)
	{
		int mid = (L+R) >> 1;
		build_tree(L,mid,o*2);
		build_tree(mid+1,R,o*2+1);
		push_up(o);
	}
}

inline void update1(int QL,int QR,SgTreeDataType v,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR)
    {
        tree[o].x += v * cos(tree[o].ang);
        tree[o].y += v * sin(tree[o].ang);
        return;
    }
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		if (QL <= mid) update1(QL,QR,v,o*2);
		if (QR >  mid) update1(QL,QR,v,o*2+1);
		push_up(o);
	}
}

inline void update2(int QL,int QR,SgTreeDataType v,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR)
    {
        double l = sqrt((tree[o].x*tree[o].x)+(tree[o].y*tree[o].y));
        tree[o].ang -= v*pi/180.0;
        tree[o].x = l * cos(tree[o].ang);
        tree[o].y = l * sin(tree[o].ang);
        return;
    }
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		if (QL <= mid) update2(QL,QR,v,o*2);
		if (QR >  mid) update2(QL,QR,v,o*2+1);
		push_up(o);
	}
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    build_tree(1,n,1);
    for(int i=1;i<=m;i++)
    {
        int op,x;
        double y;
        scanf("%d%d%lf",&op,&x,&y);
        if(op==1)update1(x,x,y,1);
        if(op==2)update2(x,x,y,1);
        printf("%.12f %.12f\n",tree[1].x,tree[1].y);
    }
    return 0;
}