二分查找：hdu 2199 Can you solve this equation?

最新推荐文章于 2023-01-11 18:36:19 发布

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最新推荐文章于 2023-01-11 18:36:19 发布

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分类专栏：基本算法文章标签：折半查找二分查找

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本文介绍了如何通过二分法在指定范围内求解一个包含四次项、三次项、二次项、一次项及常数项的多项式方程的根，特别关注方程在0到100之间的解。通过给出输入输出样例和代码实现，详细阐述了解题思路和步骤。

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Can you solve this equation?

Problem Description

Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

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 a real number Y (fabs(Y) <= 1e10);

Output

For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input

2
100
-4

Sample Output

1.6152
No solution!

解题思路：

1.建立函数f(x)=8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 - Y,将问题转换为在区间[0,100]上用二分法求方程f(x)=0的根，注意控制精度exp

2.二分思想的应用

</pre><pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <cstring>
#include<cmath>
#define maxn 100010
#define eps 1e-6
using namespace std;
double y;
double f(double x){return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6-y;}
double bsearch()
{
    double x=0,y=100;
    double m;
    while(y-x>=eps){
        m=(y+x)/2;
        if(f(m)>0) y=m;else x=m;
    }
    return m;
}
int main()
{
   // freopen("input.txt","r",stdin);
   int T;cin>>T;
   while(T--){
    cin>>y;
    if(f(0)*f(100)>0)  cout<<"No solution!"<<endl;
    else{
       printf("%.4lf\n",bsearch());
    }
   }
    return 0;
}

新手上路，望高手不吝赐教，谢谢！~