Can you solve this equation?

本文介绍了一种使用二分法求解特定多项式方程在0到100区间内的实数根的方法,并提供了完整的C++代码实现。该算法能够准确地找到方程的解或判断在指定范围内是否存在解。
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!
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
double l,h,mid,y;
double run(double x)
{
    return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>y;
        l=0;h=100;
        while(l+1e-8<h)
        {
            mid=(l+h)/2;
            if(run(mid)>y) h=mid;
            else l=mid;
        }
        if (l==0||h==100) cout<<"No solution!"<<endl;
        else printf("%.4lf\n",l);
    }
    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值