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最新推荐文章于 2024-11-01 15:10:34 发布

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最新推荐文章于 2024-11-01 15:10:34 发布

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本文介绍了一种通过二分查找法来求解特定多项式函数在闭区间[0,100]内的极小值的方法。针对给定的函数F(x)=6*x^7+8*x^6+7*x^3+5*x^2-y*x，通过编写C++程序实现精确计算，并给出了样例输入输出。

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题目：

Strange fuction

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 144 Accepted Submission(s) : 72

Problem Description

Now, here is a fuction:<br>  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)<br>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<br>100<br>200

Sample Output

-74.4291<br>-178.8534

Author

Redow

代码：#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
double f(double x,long long y)
{
return (42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x-y);
}
int main()
{
int t;
cin>>t;
while(t--)
{
double mid=50.0,m=0.0,n=100.0,x,s;
long long y;
cin>>y;
while(fabs(f(mid,y))>1e-5)
{
if((f(mid,y)>0))
{
n=mid;
mid=(m+n)/2;
}
else if(f(mid,y)<0)
{
m=mid;
mid=(m+n)/2;
}
}
x=mid;
s=6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
printf("%.4lf\n",s);
}
return 0;
}