HDU - 2199 Can you solve this equation?

最新推荐文章于 2021-01-14 20:42:38 发布

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本文介绍了一种通过二分查找法求解特定多项式方程的数值解的方法。该方程形式为8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6=Y，在0到100区间内寻找方程的根。文章提供了一个C++程序示例，用于解决此类问题，并详细说明了实现细节。

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20612 Accepted Submission(s): 9045

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!

很简单的题死活WA了好几次……首先是输入的x可能为浮点型，其次是函数是单调递增且有取值范围，超过最值则输出没有结果，最后是计算精度，虽然是保留4位但是还是要计算到1e-8以内……

#include <iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

int main()
{
    int T;
    double x,y,z,l,h;
    cin>>T;
    while(T--){

        cin>>y;
        if(y<6||y>807020306){
            cout<<"No solution!"<<endl;
            continue;
        }
        x=50.0;
        h=100.0;
        l=0.0;
        while(fabs(h-l)>1e-8){
            z=8.0*x*x*x*x + 7.0*x*x*x + 2.0*x*x + 3.0*x + 6.0;
            if(z>y){
                h=x;
            }else{
                l=x;
            }
            x=(h+l)/2;
        }
        //printf("%.5lf\n",x);
        printf("%.4lf\n",x);

    }
    return 0;
}