Equation Again 最大化最小值

最新推荐文章于 2022-07-26 00:00:18 发布

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最新推荐文章于 2022-07-26 00:00:18 发布

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原文链接：http://www.cnblogs.com/cyd308/p/4681296.html

本文探讨了一种用于解决特定形式方程的算法，该算法通过数学优化和搜索技术，实现了对未知数的准确求解。对于每条输入数据，程序会计算并输出精确到小数点后五位的解，若无解则提示特定信息。

C - Equation Again

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

This problem’s author is too lazy to write the problem description, so he only give you a equation like X ^(eY) == (eY) ^x, and the value of Y, your task is calculate the value of X.
Note : here e is the Natural logarithm.

Input

Each line will contain one number Y(Y >= 1). Process to end of file.

Output

For each case, output X on one line, accurate to five decimal places, if there are many answers, output them in increasing order, if there is no answer, just output “Happy to Women’s day!”.

Sample Input

Sample Output

2.71828

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 const double INF=0x3f3f3f3f;
 5 const double s=1e-9;
 6 const double e=exp(1);
 7 double y,Y;
 8 
 9 bool C(double x)
10 {
11     if((log(x)/x)<(log(e*y)/(e*y)))
12         return true;
13     else
14         return false;
15 }
16 
17 int main()
18 {
19     int i,j;
20     while(scanf("%lf",&y)!=EOF)
21     {
22         double Y=(log(e*y)/(e*y));
23         if(y==1.0)
24             printf("%.5lf\n",e);
25         else if(Y>(log(e)/e))
26         {
27             printf("Happy to Women’s day!\n");
28         }
29         else
30         {
31             double lb=1.0,ub=e;
32             for(i=1;i<=1000;i++)
33             {
34                 double mid=(lb+ub)/2;
35                 if(C(mid))
36                     lb=mid;
37                 else
38                     ub=mid;
39             }
40             printf("%.5lf ",ub);
41             lb=e,ub=INF;
42             for(i=1;i<=1000;i++)
43             {
44                 double mid=(lb+ub)/2;
45                 if(C(mid))
46                     ub=mid;
47                 else
48                     lb=mid;
49             }
50             printf("%.5lf\n",lb);
51 
52         }
53     }
54     return 0;
55 }