题目描述

Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 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!

大意

求解方程的简单水题

思路

简单的水题，使用二分法即可。但需要注意浮点数比较时的精度，我写的是1e-5，如果精度太高的话会超时的

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<iomanip>
using namespace std;
double m, res = 0;
int ji = 0;
double f(double res)
{
    return res*res*res*res * 8 + res*res*res * 7 + res*res * 2 + res * 3 + 6;
}
int main()
{
    //cin.sync_with_stdio(false);
    //freopen("date.in", "r", stdin);
    //freopen("date.out", "w", stdout);
    int N, temp;
    double b, e, tem = 50;
    scanf("%d", &N);
    for (int i = 0; i<N; i++){
        b = 0, e = 100, tem = 50;
        cin >> m;
        if (m<6 || m>807020306)
            //cout << "No solution!" << endl;
            printf("No solution!\n");
        else{
            while (fabs(f(tem) - m) >= 1e-5)
                if (f(tem)>m){
                    e = tem;
                    tem = (b + e) / 2;
                }
                else{
                    b = tem;
                    tem = (b + e) / 2;
                }
                //cout << fixed << setprecision(4) << tem << endl;
                 printf("%0.4f\n", tem);
        }
    }

}