SGU 403 Game with points

408. Game with points

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

Recently Petya has discovered new game with points. Rules of the game are quite simple. First, there is only one point A₀ with coordinates (0, 0). Then Petya have to draw N another points. Points must be drawn consequently and each new point must be connected with exactly one of the previous points by a segment. Let's decribe the game process more formally. At the i-th step Petya chooses the position of the point A_i (not necessarily with integer coordinates). Than he chooses one of the previously drawn points in order to connect it with the point A_i. Lets call this point B. The following conditions must be held:

• Point A_i must not coincide with any of the previous points.
• Point A_i must not lie on the previously drawn segments.
• Segment A_iB must not have common points with previously drawn segments, except possibly the point B.
• Segment A_iB must not cover any of the previous points, except the point B.
• Length of the segment A_iB must not exceed 1. After drawing each point Petya computes two values.
• The largest number of segments which share a common point.
• The largest euclid distance between some pair of points. After each step Petya gains the score which is equal to the product of these values. Find out which is the maximal score Petya can gain after the whole game.
  
  Input
  Input contains single integer number N (0 ≤ N ≤ 1000).
  
  Output
  Output the maximal score that Petya can gain. Your answer must be accurate up to 10^-3.
  
  Example(s)

  sample input	sample output
  2	5.000

  
  

  sample input

  sample output

  4

  20.000 #include<iostream> #include<string.h> #include<stdio.h> #include<ctype.h> #include<algorithm> #include<stack> #include<queue> #include<set> #include<math.h> #include<vector> #include<map> #include<deque> #include<list> using namespace std; int a[1111]; int b[1111]; int main() { //freopen("in.txt", "r", stdin); int n; scanf("%d", &n); a[1] = 1; b[1] = 1; if (n == 0) printf("0.000 "); else if (n == 1)printf("1.000 "); else if (n == 2)printf("5.000 "); else { double ans = 5; int m = 2, d = 2; for (int i = 2; i <n; i++) { if ((m + 1)*d >= m*(d + 1)) m++; else d++; ans += m*d; } printf("%.3lf ", ans); } return 0; }