二分搜索 HDOJ 2675 Equation Again

 

题目传送门

 1 /*
 2     二分搜索：式子两边取对数，将x提出来，那么另一边就是一个常数了，函数是：lnx/x。二分搜索x，注意要两次 
 3 */
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cmath>
 7 using namespace std;
 8 
 9 const double e = exp (1.0);
10 
11 double cal(double x) {
12     return log (x) / x;
13 }
14 
15 int main(void)  {       //HDOJ 2675 Equation Again
16     //freopen ("HDOJ_2675.in", "r", stdin);
17 
18     double y, k;
19     while (scanf ("%lf", &y) == 1)  {
20         k = (1 + log (y)) / (e * y);
21         //if (k > 1 / e)  {                 //囧！
22         //    puts ("Happy to Women’s day!");    continue;
23         //}
24         if (k == 1 / e) {
25             printf ("%.5f\n", e); continue;
26         }
27 
28         double x1, x2;
29         double l = 0, r = e;
30         for (int i=1; i<=100; ++i)  {
31             double mid = (l + r) / 2;
32             if (cal (mid) >= k)   r = mid;
33             else    l = mid;
34         }
35         x1 = l;
36         l = e, r = 1e9;
37         for (int i=1; i<=100; ++i)  {
38             double mid = (l + r) / 2;
39             if (cal (mid) >= k)   l = mid;
40             else    r = mid;
41         }
42         x2 = l;
43         printf ("%.5f %.5f\n", x1, x2);
44     }
45 
46     return 0;
47 }

 

posted @ 2015-07-25 18:23  Running_Time  阅读(181)  评论(0)  编辑  收藏  举报