• 杭电2899--Strange fuction(二、三分)


    Strange fuction

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4603    Accepted Submission(s): 3308


    Problem Description
    Now, here is a fuction:
      F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
    Can you find the minimum value when x is between 0 and 100.
     

     

    Input
    The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
     

     

    Output
    Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
     

     

    Sample Input
    2 100 200
     

     

    Sample Output
    -74.4291 -178.8534
     

     

    Author
    Redow
     

     

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    lcy   |   We have carefully selected several similar problems for you:  2289 2298 3400 2446 1399 
    RE: 求最值问题。
    //三分法:
     1 #include <cstdio>
     2 #include <iostream>
     3 using namespace std;
     4 double f(double x, double y)
     5 {
     6     return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
     7 }
     8 int main()
     9 {
    10     int t; double m;
    11     scanf("%d", &t);
    12     while(t--)
    13     {
    14         scanf("%lf", &m);
    15         double l_min, r_max, min = 0.0, max = 100.0;
    16     //    printf("%.4lf %.4lf
    ", min, max);
    17         while(max - min > 1e-8)
    18         {
    19             l_min = (2 * min + max) / 3.0;
    20             r_max = (min + 2 * max) / 3.0;
    21             if(f(l_min, m)>f(r_max, m)) min = l_min;
    22             else max = r_max; 
    23         }
    24     //    printf("%.4lf %.4lf
    ", min, max);
    25         printf("%.4lf
    ",f((min + max) / 2.0, m));
    26     }
    27     return 0;
    28 } 

    //闲来无事, 敲敲手生的二分(大同小异, 一个比较值, 一个比较导数);

     1 #include <cstdio>
     2 #include <iostream>
     3 using namespace std;
     4 double f(double x, double y)
     5 {
     6     return 42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x-y;
     7 }
     8 double F(double x, double y)
     9 {
    10     return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x; 
    11 }
    12 int main()
    13 {
    14     int t;
    15     scanf("%d", &t);
    16     while(t--)
    17     {
    18         double m, min = 0.0, max = 100.0, mid;
    19         scanf("%lf", &m);
    20         while(max - min > 1e-8)
    21         {
    22             mid = (max + min) / 2.0;
    23             if(f(mid, m) <= 0) min = mid;
    24             else max = mid;
    25         }
    26         printf("%.4lf
    ", F((min + max) / 2.0 , m));
    27     }
    28     return 0;
    29 } 
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  • 原文地址:https://www.cnblogs.com/soTired/p/4701939.html
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