Description
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., 180o. The operator issues commands that change the angle in exactly one joint.
Input
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 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
Source
题解：
我是没想到是线段树，看了是线段树后自己yy了一个建线段树的方法，好像跟书上的不一样。
书上的方法：
我的代码：
program rrr(input,output);
const
eps=1e-10;
type
treetype=record
l,r:longint;
x,y,d:double;
end;
var
a:array[..]of treetype;
c:array[..]of double;
b:array[..]of longint;
n,m,i,x:longint;
y,t,xx,yy:double;
procedure build(k,l,r:longint);
var
mid,i:longint;
begin
a[k].l:=l;a[k].r:=r;a[k].x:=;a[k].d:=;
if l=r then begin a[k].y:=b[l];exit; end;
mid:=(l+r)>>;i:=k+k;
build(i,l,mid);build(i+,mid+,r);
a[k].y:=a[i].y+a[i+].y;
end;
procedure pushdown(k:longint);
var
i:longint;
begin
if a[k].l=a[k].r then a[k].d:=;
if abs(a[k].d)<eps then exit;
i:=k+k;
xx:=a[i].x*cos(a[k].d)-a[i].y*sin(a[k].d);
yy:=a[i].x*sin(a[k].d)+a[i].y*cos(a[k].d);
a[i].x:=xx;a[i].y:=yy;a[i].d:=a[i].d+a[k].d;
inc(i);
xx:=a[i].x*cos(a[k].d)-a[i].y*sin(a[k].d);
yy:=a[i].x*sin(a[k].d)+a[i].y*cos(a[k].d);
a[i].x:=xx;a[i].y:=yy;a[i].d:=a[i].d+a[k].d;
a[k].d:=;
end;
procedure change(k:longint);
var
mid,i:longint;
begin
pushdown(k);
if x<a[k].l then
begin
xx:=a[k].x*cos(t)-a[k].y*sin(t);
yy:=a[k].x*sin(t)+a[k].y*cos(t);
a[k].x:=xx;a[k].y:=yy;
a[k].d:=t;
exit;
end;
mid:=(a[k].l+a[k].r)>>;i:=k+k;
if x<mid then change(i);
change(i+);
a[k].x:=a[i].x+a[i+].x;a[k].y:=a[i].y+a[i+].y;
end;
begin
assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
while not eof do
begin
readln(n,m);if (n=) and (m=) then break;
for i:= to n do begin read(b[i]);c[i]:=pi; end;
build(,,n);
for i:= to m do
begin
readln(x,y);t:=y/*pi-c[x];c[x]:=y/*pi;
change();
writeln(a[].x::,' ',a[].y::);
end;
writeln;
end;
close(input);close(output);
end.