Sgu 120

120. Archipelago

time limit per test: 0.25 sec. memory limit per test: 4096 KB

Archipelago Ber-Islands consists of N islands that are vertices of equiangular and equilateral N-gon. Islands are clockwise numerated. Coordinates of island N₁ are (x₁, y₁), and island N₂ – (x₂, y₂). Your task is to find coordinates of all N islands.

Input

In the first line of input there are N, N₁ and N₂ (3£ N£ 150, 1£ N₁,N₂£N, N₁¹N₂) separated by spaces. On the next two lines of input there are coordinates of island N₁ and N₂ (one pair per line) with accuracy 4 digits after decimal point. Each coordinate is more than -2000000 and less than 2000000.

Output

Write N lines with coordinates for every island. Write coordinates in order of island numeration. Write answer with 6 digits after decimal point.

Sample Input

4 1 3
1.0000 0.0000
1.0000 2.0000

Sample Output

1.000000 0.000000
0.000000 1.000000
1.000000 2.000000
2.000000 1.000000

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <set>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff 
12 #define mod 1000000007
13 #define ll long long
14 #define maxn 2005
15 #define pi acos(-1.0)
16 #define dis(a, b) sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y))
17 ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
18 int n, m, k, c, a, b;    
19 struct point{
20     double x, y;
21 }p[maxn];
22 int main(){
23     double tmp, r, t;
24     scanf("%d%d%d", &n,&a, &b);
25     scanf("%lf%lf%lf%lf", &p[a].x,&p[a].y,&p[b].x,&p[b].y);
26     if (a > b)swap(a, b);
27     r = dis(p[a], p[b]) / sin(pi * (b - a) / n) / 2;
28     point o;
29     o.x = (p[a].x + p[b].x) / 2 + (p[b].y - p[a].y) / tan(pi * (b - a) / n) / 2;
30     o.y = (p[a].y + p[b].y) / 2 - (p[b].x - p[a].x) / tan(pi * (b - a) / n) / 2;
31     tmp = asin((p[a].y - o.y)/r);
32     if (p[a].x<o.x)tmp = pi - tmp;
33     for (int i = 1; i <= n; i++){
34         t = tmp + 2 * pi*(a - i) / n;
35         p[i].x = o.x + r*cos(t);
36         p[i].y = o.y + r*sin(t);
37     }
38     for (int i = 1; i <= n; i++)printf("%.6lf %.6lf\n", p[i].x, p[i].y);
39     return 0;
40 }

View Code

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

posted @ 2014-02-10 04:34 HaibaraAi 阅读(136) 评论(0) 收藏举报

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