Codeforces Round #478 (Div. 2) E. Hag's Khashba

E. Hag's Khashba

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hag is a very talented person. He has always had an artist inside him but his father forced him to study mechanical engineering.

Yesterday he spent all of his time cutting a giant piece of wood trying to make it look like a goose. Anyway, his dad found out that he was doing arts rather than studying mechanics and other boring subjects. He confronted Hag with the fact that he is a spoiled son that does not care about his future, and if he continues to do arts he will cut his 25 Lira monthly allowance.

Hag is trying to prove to his dad that the wooden piece is a project for mechanics subject. He also told his dad that the wooden piece is a strictly convex polygon with $\mathrm{nn vertices.}$

Hag brought two pins and pinned the polygon with them in the $\mathrm{11-st\; and 22-nd\; vertices\; to\; the\; wall.\; His\; dad\; has qq queries\; to\; Hag\; of\; two\; types.}$

• $\mathrm{11 ff tt:\; pull\; a\; pin\; from\; the\; vertex ff,\; wait\; for\; the\; wooden\; polygon\; to\; rotate\; under\; the\; gravity\; force\; (if\; it\; will\; rotate)\; and\; stabilize.\; And\; then\; put\; the\; pin\; in\; vertex tt.}$
• $\mathrm{22 vv:\; answer\; what\; are\; the\; coordinates\; of\; the\; vertex vv.}$

Please help Hag to answer his father's queries.

You can assume that the wood that forms the polygon has uniform density and the polygon has a positive thickness, same in all points. After every query of the 1-st type Hag's dad tries to move the polygon a bit and watches it stabilize again.

Input

The first line contains two integers $\mathrm{nn and qq (3\le n\le 100003\le n\le 10000, 1\le q\le 2000001\le q\le 200000) —\; the\; number\; of\; vertices\; in\; the\; polygon\; and\; the\; number\; of\; queries.}$

The next $\mathrm{nn lines\; describe\; the\; wooden\; polygon,\; the ii-th\; line\; contains\; two\; integers xixi and yiyi (|xi|,|yi|\le 108|xi|,|yi|\le 108) —\; the\; coordinates\; of\; the ii-th\; vertex\; of\; the\; polygon.\; It\; is\; guaranteed\; that\; polygon\; is\; strictly\; convex\; and\; the\; vertices\; are\; given\; in\; the\; counter-clockwise\; order\; and\; all\; vertices\; are\; distinct.}$

The next $\mathrm{qq lines\; describe\; the\; queries,\; one\; per\; line.\; Each\; query\; starts\; with\; its\; type 11 or 22.\; Each\; query\; of\; the\; first\; type\; continues\; with\; two\; integers ff and tt (1\le f,t\le n1\le f,t\le n) —\; the\; vertex\; the\; pin\; is\; taken\; from,\; and\; the\; vertex\; the\; pin\; is\; put\; to\; and\; the\; polygon\; finishes\; rotating.\; It\; is\; guaranteed\; that\; the\; vertex ff contains\; a\; pin.\; Each\; query\; of\; the\; second\; type\; continues\; with\; a\; single\; integer vv (1\le v\le n1\le v\le n) —\; the\; vertex\; the\; coordinates\; of\; which\; Hag\; should\; tell\; his\; father.}$

It is guaranteed that there is at least one query of the second type.

Output

The output should contain the answer to each query of second type — two numbers in a separate line. Your answer is considered correct, if its absolute or relative error does not exceed ${\mathrm{10-410-4.}}^{}$

Formally, let your answer be $\mathrm{aa,\; and\; the\; jury\text{'}s\; answer\; be bb.\; Your\; answer\; is\; considered\; correct\; if |a-b|max(1,|b|)\le 10-4|a-b|max(1,|b|)\le 10-4}$

Examples

input

Copy

3 4
0 0
2 0
2 2
1 1 2
2 1
2 2
2 3

output

Copy

3.4142135624 -1.4142135624
2.0000000000 0.0000000000
0.5857864376 -1.4142135624

input

Copy

3 2
-1 1
0 0
1 1
1 1 2
2 1

output

Copy

1.0000000000 -1.0000000000

Note

In the first test note the initial and the final state of the wooden polygon.

Red Triangle is the initial state and the green one is the triangle after rotation around $(2,0)(2,0).$

In the second sample note that the polygon rotates $\mathrm{180180 degrees\; counter-clockwise\; or\; clockwise\; direction\; (it\; does\; not\; matter),\; because\; Hag\text{'}s\; father\; makes\; sure\; that\; the\; polygon\; is\; stable\; and\; his\; son\; does\; not\; trick\; him.}$

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const long double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const long double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, q;
 45 struct point {
 46     long double x, y;
 47     point() {}
 48     point (long double x, long double y) : x(x), y(y) {}
 49     void input() {
 50         //scanf("%lf%lf", &x, &y);
 51         cin >> x >> y;
 52     }
 53     void output() {
 54         double tx = x, ty = y;
 55         printf("%.12f %.12f\n", tx, ty);
 56     }
 57     point operator - (const point &p) const {
 58         return point(x - p.x, y - p.y);
 59     }
 60     point operator + (const point &p) const {
 61         return point(x + p.x, y + p.y);
 62     }
 63     long double operator ^ (const point &p) const {
 64         return x * p.y - y * p.x;
 65     }
 66     point operator / (const long double &k) const {
 67         return point(x / k, y / k);
 68     }
 69     point operator * (const long double &k) const {
 70         return point(x * k, y * k);
 71     }
 72     long double dis(const point &p) const {
 73         return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
 74     }
 75 } p[100015];
 76 long double d0[100015], d1[100015], th[100015];
 77 long double cal_th(point p1, point p0, point p2) {
 78     long double th1 = atan2(p1.y - p0.y, p1.x - p0.x);
 79     long double th2 = atan2(p2.y - p0.y, p2.x - p0.x);
 80     long double res = th1 - th2;
 81     if (res < 0) res += 2 * pi;
 82     return res;
 83 }
 84 multiset<int> rem;
 85 point get_pos(int index) {
 86     long double th1 = atan2(p[1].y - p[0].y, p[1].x - p[0].x);
 87     long double th2 = th1 - th[index];
 88     if (th2 < 0) th2 += 2 * pi;
 89     return p[0] + point(d0[index] * cos(th2), d0[index] * sin(th2));
 90 }
 91 void rot_and_update(int index2) {
 92     point p2 = get_pos(index2);
 93     long double th01 = atan2(p[0].y - p2.y, p[0].x - p2.x);
 94     long double th02 = 3 * pi / 2;
 95     long double th0 = th02 - th01;
 96     if (th0 < 0) th0 += 2 * pi;
 97     long double th11 = atan2(p[1].y - p2.y, p[1].x - p2.x);
 98     //p[0].output(), p[1].output(), p2.output();
 99     long double th12 = th11 + th0;
100     if (2 * pi <= th12) th12 -= 2 * pi;
101     //printf("index2 = %d\n", index2);
102     p[0] = point(p2.x, p2.y - d0[index2]);
103     p[1] = p2 + point(d1[index2] * cos(th12), d1[index2] * sin(th12));
104 }
105 void get_wp() {
106     long double s = 0;
107     p[0] = point(0, 0);
108     for (int i = 2; i < n; ++i) {
109         long double ts = (p[i] - p[1]) ^ (p[i] - p[i + 1]);
110         p[0] = p[0] + (p[1] + p[i] + p[i + 1]) / 3 * ts;
111         s += ts;
112     }
113     p[0] = p[0] / s;
114 }
115 int main() {
116     scanf("%d%d", &n, &q);
117     p[0] = point(0, 0);
118     for (int i = 1; i <= n; ++i) {
119         p[i].input();
120     }
121     get_wp();
122     for (int i = 1; i <= n; ++i) {
123         th[i] = cal_th(p[1], p[0], p[i]);
124         d0[i] = p[0].dis(p[i]);
125         d1[i] = p[1].dis(p[i]);
126     }
127     rem.insert(1), rem.insert(2);
128     while (q--) {
129         int op;
130         scanf("%d", &op);
131         if (1 == op) {
132             int f, t;
133             scanf("%d%d", &f, &t);
134             multiset<int>::iterator it = rem.lower_bound(f);
135             rem.erase(it);
136             int v = *rem.begin();
137             rot_and_update(v);
138             rem.insert(t);
139         } else {
140             int v;
141             scanf("%d", &v);
142             get_pos(v).output();
143         }
144     }
145     return 0;
146 }
147 /*
148 10 10
149 0 -100000000
150 1 -100000000
151 1566 -99999999
152 2088 -99999997
153 2610 -99999994
154 3132 -99999990
155 3654 -99999985
156 4176 -99999979
157 4698 -99999972
158 5220 -99999964
159 1 2 5
160 2 1
161 1 1 7
162 2 5
163 1 5 4
164 1 4 2
165 2 8
166 1 7 9
167 2 1
168 1 2 10
169 */

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