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  • 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
    ", 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
    ", x1, x2);
    44     }
    45 
    46     return 0;
    47 }
    编译人生,运行世界!
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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4676351.html
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