POJ2991:Crane(线段树+几何)

Description

ACM has bought a new crane (crane -- jeřáb) . The crane consists of n segments of various lengths, connected by flexible joints. The end of the i-th segment is joined to the beginning of the i + 1-th one, for 1 ≤ i < n. The beginning of the first segment is fixed at point with coordinates (0, 0) and its end at point with coordinates (0, w), where w is the length of the first segment. All of the segments lie always in one plane, and the joints allow arbitrary rotation in that plane. After series of unpleasant accidents, it was decided that software that controls the crane must contain a piece of code that constantly checks the position of the end of crane, and stops the crane if a collision should happen.

Your task is to write a part of this software that determines the position of the end of the n-th segment after each command. The state of the crane is determined by the angles between consecutive segments. Initially, all of the angles are straight, i.e., 180^o. The operator issues commands that change the angle in exactly one joint.

Input

The input consists of several instances, separated by single empty lines.

The first line of each instance consists of two integers 1 ≤ n ≤10 000 and c 0 separated by a single space -- the number of segments of the crane and the number of commands. The second line consists of n integers l1,..., ln (1 li 100) separated by single spaces. The length of the i-th segment of the crane is li. The following c lines specify the commands of the operator. Each line describing the command consists of two integers s and a (1 ≤ s < n, 0 ≤ a ≤ 359) separated by a single space -- the order to change the angle between the s-th and the s + 1-th segment to a degrees (the angle is measured counterclockwise from the s-th to the s + 1-th segment).

Output

The output for each instance consists of c lines. The i-th of the lines consists of two rational numbers x and y separated by a single space -- the coordinates of the end of the n-th segment after the i-th command, rounded to two digits after the decimal point.

The outputs for each two consecutive instances must be separated by a single empty line.

Sample Input

2 1
10 5
1 90

3 2
5 5 5
1 270
2 90

Sample Output

5.00 10.00

-10.00 5.00
-5.00 10.00

题意：一开始有n根木棍，木棍都在y轴上并有自己的长度，有m次操作，每次操作将i和i+1木棍的夹角调成给出的角度，每次调整输出木棍末端的坐标

思路：这道题也就是一道线段树的更新问题，要注意的也就是角度的变化与坐标的变化，要注意的是，这道题输出用%.2lf的话C++能过但G++超时，但用%.2f输出就不超时了，其中原理我也不是太清楚

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define lson i<<1
#define rson i<<1|1
#define lc l,mid,i<<1
#define rc mid+1,r,i<<1|1
const int L = 100000+10;
const double pi = acos(-1.0);

struct node
{
    double x,y;
    int deg;
    int flag;
} a[L<<2];

double set(int x)
{
    return x*pi/180;
}

void work(int i,int deg)//求新坐标公式
{
    double r = set(deg);
    double x = a[i].x;
    double y = a[i].y;
    a[i].x = x*cos(r)-y*sin(r);
    a[i].y = x*sin(r)+y*cos(r);
    a[i].deg = (a[i].deg+deg)%360;
}

void pushup(int i)
{
    a[i].x = a[lson].x+a[rson].x;
    a[i].y = a[lson].y+a[rson].y;
}

void pushdown(int i)
{
    if(a[i].flag)
    {
        work(lson,a[i].flag);
        work(rson,a[i].flag);
        a[lson].flag+=a[i].flag;
        a[rson].flag+=a[i].flag;
        a[i].flag = 0;
    }
}

void init(int l,int r,int i)
{
    a[i].x = a[i].y = 0;
    a[i].flag = a[i].deg = 0;
    if(l == r)
    {
        scanf("%lf",&a[i].y);
        return;
    }
    int mid = (l+r)>>1;
    init(lc);
    init(rc);
    pushup(i);
}

void insert(int l,int r,int i,int L,int R,int z)
{
    if(L<=l && r<=R)
    {
        work(i,z);
        a[i].flag+=z;
        return ;
    }
    pushdown(i);
    int mid = (l+r)>>1;
    if(L<=mid)
        insert(lc,L,R,z);
    if(R>mid)
        insert(rc,L,R,z);
    pushup(i);
}

int query(int l,int r,int i,int x)
{
    if(l == r)
        return a[i].deg;
    pushdown(i);
    int mid = (l+r)>>1;
    if(x<=mid)
        return query(lc,x);
    else
        return query(rc,x);
}
int main()
{
    int n,m,x,y,flag = 1,i,j;
    while(~scanf("%d%d",&n,&m))
    {
        init(0,n-1,1);
        if(!flag)
            printf("\n");
        flag = 0;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            int deg;
            deg  = query(0,n-1,1,x-1)+180+y-query(0,n-1,1,x);//由于题目是逆时针转的，我的计算是顺时针，要加上180度,将后面的棒看成依然在Y轴，所以要减去后一个的角度
            insert(0,n-1,1,x,n-1,deg);
            printf("%.2f %.2f\n",a[1].x,a[1].y);
        }
    }

    return 0;
}