Can you solve this equation?(二分）

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!

 

题意：

x∈[0, 100], Y∈[-10^10, 10^10]，求8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = Y 的解。

 

思路：

通过求导发现，f(x)是一个单调递增函数。即随着x的增大，y逐渐增大。成一个正比例的关系。所以可以采用二分法来解决此题。

注意：此题要求x精确到4位，我以为是要以x取到4位小数为结束条件，后来看了别人代码后发现是错的。此题应该以y1，y2很接近作为结束条件。原因：当x取到4位时，在4次方什么的会使y仍然很不精确。所以，直接索性让y取到4位保证更精确。如果讨论x的话，则应 取精确度更高的。

代码：

#include <cstdio>
#include <cmath>
double f (double x)  
{
    return 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6;
}
int main()
{
    int T;
    double x1, x2, x3, y, y1, y2, y3;
    scanf("%d", &T);
    while(T--)
    {
        x1 = 0;
        x2 = 100;        
        scanf("%lf", &y);  
        y1 = f(x1) - y;
        y2 = f(x2) - y;
        if (y1 > 0 || y2 < 0)
            printf("No solution!\n");
        else
        {
            while (fabs(y1 - y2) >= 0.0001)
            {
                x3 = (x1 + x2) / 2;
                y3 = f(x3) - y;
                if (y3 >= 0)
                    x2 = x3;
                else
                    x1 = x3;
                y1 = f(x1) - y;
                y2 = f(x2) - y;
            }
            printf("%0.4f\n", x3);
        }
    }
    return 0;
}