Can you solve this equation?



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!
   
   
   
   

思路：题目中的关于x的式子是单调递增的，当<f(0) ||  >f(100)时不存在，所以可以采用二分法不断缩小范围，但要注意二分法的终止条件

#include <stdio.h>
#include <math.h>
double f(double x)
{
    return (8.0*x*x*x*x+7.0*x*x*x+2*x*x+3*x+6);
}

int main()
{
    //freopen("e:\\in.txt","r",stdin);
     int T;
     double re;
    scanf("%d",&T);
    while(T--)
    {
       scanf("%lf",&re);
       if(f(0)>re||f(100)<re)
        printf("No solution!\n");
       else
       {
           double low=0.0,high=100.0,middle=50.0;
           while(fabs(f(middle)-re)>1e-5)
           {
               if(f(middle)>re)
                  high=middle;
               else if(f(middle)<re)
                   low=middle;

                middle=(low+high)/2;
           }
           printf("%.4lf\n",middle);

       }
    }
    return 0;
}

要点：1.二分法的终止条件：fabs（）<1e-5,意义为当其小于0时，（二分法的终止条件要根据题目来判断）

2.从键盘上输入的是整数，但如果它的类型是double，则实际是浮点型数字。