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 Ai (not necessarily with integer coordinates). Than he chooses one of the previously drawn points in order to connect it with the point Ai. Lets call this point B. The following conditions must be held:

  • Point Ai must not coincide with any of the previous points.
  • Point Ai must not lie on the previously drawn segments.
  • Segment AiB must not have common points with previously drawn segments, except possibly the point B.
  • Segment AiB must not cover any of the previous points, except the point B.
  • Length of the segment AiB 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
     1 #include<iostream>  
 2 #include<string.h>  
 3 #include<stdio.h>  
 4 #include<ctype.h>  
 5 #include<algorithm>  
 6 #include<stack>  
 7 #include<queue>  
 8 #include<set>  
 9 #include<math.h>  
10 #include<vector>  
11 #include<map>  
12 #include<deque>  
13 #include<list>  
14 using namespace std;
15 int a[1111];
16 int b[1111];
17 
18 int main()
19 {
20     //freopen("in.txt", "r", stdin);
21     int n;
22     scanf("%d", &n);
23     a[1] = 1;
24     b[1] = 1;
25     
26     if (n == 0) printf("0.000\n");
27     else if (n == 1)printf("1.000\n");
28     else if (n == 2)printf("5.000\n");
29     else
30     {
31         double ans = 5;
32         int m = 2, d = 2;
33         for (int i = 2; i <n; i++)
34         {
35             if ((m + 1)*d >= m*(d + 1))
36                 m++;
37             else
38                 d++;
39             ans += m*d;
40         }
41         printf("%.3lf\n", ans);
42     }
43 
44     return 0;
45 }
    View Code

     
posted @ 2014-07-31 09:52  qscqesze  阅读(205)  评论(0)  编辑  收藏  举报