HDU 2899 Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 778    Accepted Submission(s): 597

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

 

 

 

题意是求最小值。

和HDU2199差不多。。。。。那里是原方程的解。。。。这里先求下导。。。。。然后找到极小值点就OK。。。。

 

 

 

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cctype>
#include <cstring>
#include <sstream>
#include <fstream>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <algorithm>

using namespace std;
//Constant Declaration
/*--------------------------*/
//#define LL long long 
#define LL __int64
const int M=1000000;
const int INF=1<<30;
const double EPS = 1e-11;
const double PI = acos(-1.0);
/*--------------------------*/
// some essential funtion
/*----------------------------------*/
void Swap(int &a,int &b){   int t=a;a=b;b=t; }
int Max(int a,int b){   return a>b?a:b;  }
int Min(int a,int b){   return a<b?a:b;  }
int Gcd(int a,int b){   while(b){b ^= a ^=b ^= a %= b;}  return a; }
/*----------------------------------*/
//for (i = 0; i < n; i++)
/*----------------------------------*/
#define F(x) (42*pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x - y)
#define ans(x) (6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x)
double y;
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t, case1 = 0;
    scanf("%d", &t);
    int n, m;
    int i, j;
    //scanf("%d%d", &n, &m);

    while (t--)
    {
        scanf("%lf", &y);
        if (F(0.0) * F(100.0) > 0)
        {
            if (F(0.0) > 0)
            {
                printf("%.4lf\n", ans(0.0));
            }
            else
            {
                printf("%.4lf\n", ans(100.0));
            }
        }
        else
        {
            double l = 0, r = 100;
            if (F(l) == 0)
            {
                printf("%.4lf", ans(0.0));
                goto end;
            }
            if (F(r) == 0)
            {
                printf("%.4lf", ans(100.0));
                goto end;
            }

            double min;
            while (r - l > EPS)
            {
                min = (l + r) / 2;
                if (F(min) == 0)
                {
                    goto end;
                }
                if (F(l) * F(min) < 0)
                {
                    r = min;
                }
                else
                {
                    l = min;
                }     


            }
end : 
            printf("%.4lf\n", ans(min));




        }
    }

    return 0;
}