acm课程练习2--1002

题目描述

Now, here is a fuction:
  F(x) = 6 * x^7+8x^6+7x^3+5x^2-yx (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

大意

求函数在区间[0,100]的最小值

思路

对函数求导得F’(x)=42* x^6+48x^5+21x^2+10x-y

，再对导函数求导，发现导函数是单调递增的。

得到结论，函数的最小值点为

42 x^6+48x^5+21x^2+10*x=y的根。

程序应首先判断根是否在[0,100]区间内，分为三种情况讨论。

这一题的代码与第一题差别不大，是第一题的变形

AC代码

#include<iostream>
#include<iomanip>
#include<stdio.h>
#include<cmath>
using namespace std;
double f_d(double res)
{
    return res*res*res*res *res*res*42  + res*res*res*res*res*48 + res*res*21 + res * 10;
}
double f(double res,double y){
    return res*res*res*res*res*res*res*6 + res*res*res*res*res*res*8 + res*res*res*7 + res*res* 5-res*y;
}
int main(){
    //freopen("date.in", "r", stdin);
    //freopen("date.out", "w", stdout);
    int T;
    double a;
    double b,e,tem;
    cin>>T;
    for(int i=0;i<T;i++){
        cin>>a;
        if(f_d(100)<=a)
            cout<<fixed<<setprecision(4)<<f(100,a)<<endl;
        else
            if(f_d(0)>=a)
                cout<<fixed<<setprecision(4)<<f(0,a)<<endl;
            else{
                b = 0, e = 100, tem = 50;
                    while (fabs(f_d(tem) - a) >= 1e-7)
                        if (f_d(tem)>a){
                            e = tem;
                            tem = (b + e) / 2;
                        }
                        else{
                            b = tem;
                            tem = (b + e) / 2;
                        }
                    cout<<fixed<<setprecision(4)<<f(tem,a)<<endl;
                }
    }
}

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