搜索0之1001

1 题目编号：1001

2 题目内容：

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;<br>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<br>100<br>-4<br>

 

Sample Output

1.6152<br>No solution!<br>

 

Author

Redow

3 解题思路形成过程：此题为简单的二分查找题，从0到100进行计算，令100为最大值，0为最小值，分别代入计算公式，分别得到一个数值，两值进行相减，若值小于1e-6，则结果为（max+min）/2，否则，可设一中间值为（max+min）/2，并判断其与y的大小，若比y大，则max=中间值，否则，最小值=中间值，然后，进行下一轮循环，直至满足条件为止。

4 代码：

#include <iostream>
#include <cmath>
using namespace std;
double search(double x)
{
return 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6;
}
int main()
{
int n;
cin >> n;
cout.precision(4);
while (n--)
{
int y;
double first, last, mid;
cin >> y;
if (search(0) <= y && search(100) >= y)
{
first = 0;
last = 100;
while (last - first>1e-6)
{
mid = (first + last) / 2;
if (search(mid)>y)
last = mid;
else
first = mid;
}
cout << fixed << (last + last) / 2 << endl;
}
else
cout << "No solution!" << endl;
}
return 0;
}