Problem Description
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!
Author
Redow
思路
这个问题可以用二分法解决,因为给定的函数是单调的,所以可以根据要求的精度不断二分找答案
详见代码
代码
#include<bits/stdc++.h>
using namespace std;
double f(double x)
{
return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6 ;
}
int main()
{
int n;
cin >> n;
while(n--)
{
double t;
cin >> t;
double l = 0, r = 100.0;
if(t<f(0) || t>f(100.0))
{
cout << "No solution!
";
continue;
}//因为函数单调,所以可以这么判断
double mid;
while(r-l>1e-8)
{
mid = (l+r)/2;
double ans = f(mid);
if(ans>t)
r = mid;
if(ans<t)
l = mid;
}
printf("%.4lf
",l);
}
return 0;
}