POJ1064 

 1 #include <iostream>
 2 #include <iomanip> 
 3 #include <cmath>
 4 using namespace std;
 5 
 6 int N;
 7 int K;
 8 double num[10001];
 9 
10 int calc(double len)
11 {
12     int numK= 0;
13     for(int i = 0; i < N; ++i)
14     {
15         numK += (int)(num[i] / len);    
16     }
17     return numK;
18 }
19 
20 double solve()
21 {
22     double low = 0;
23     double upper = 100000;
24     int count  = 0;
25     double mid;
26     int numK;
27     while(count < 100)
28     {
29         mid = (low + upper) / 2;
30         numK = calc(mid);
31         if(numK >= K)
32             low = mid;    
33         else
34             upper = mid;
35         ++count;    
36     }
37     return mid;
38 }
39 
40 int main()
41 {
42     //基于二分法的思想实现 
43     cin >> N >> K;
44     for(int i = 0; i < N; ++i)
45         cin >> num[i];
46         //这里floor很重要 
47     cout << setiosflags(ios::fixed) << setprecision(2) << floor(solve()*100)/100 << endl;
48     return 0;
49 }

 

posted @ 2014-10-17 00:00  多解方程式  阅读(252)  评论(0)    收藏  举报