Can you solve this equation?

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)

Total Submission(s):     Accepted Submission(s): 

Problem Description

Now,given the equation *x^ + *x^ + *x^ + *x +  == Y,can you find its solution between  and ;

Now please try your lucky.

Input

The first line of the input contains an integer T(<=T<=) 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  decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between  and .

Sample Input

-

Sample Output

1.6152

No solution!

Author

Redow

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lcy
 

 #include"iostream"

 #include"algorithm"

 #include"cstdio"

 #include"cstring"

 #include"cmath"

 #define max(a,b) a>b?a:b

 #define min(a,b) a<b?a:b

 #define MX 10000 + 50

 using namespace std;

 double f(double x)

 {

     return  *pow(x,)+*pow(x,)+*pow(x,)+*x+;

 }

 int main()

 {

     int n;

     double m;

     while(~scanf("%d",&n))

     {

         for(int k=; k<=n; k++)

         {

             scanf("%lf",&m);

             double i=0.0;

             if(m<f()||m>f())

             {

                 printf("No solution!\n");continue;

             }

             double fr=0.0,ed=100.0;

             for(; fabs(f(fr)-f(ed))>1e-;)

             {

                 i=(fr+ed)/;

                 if(f(i)<m)fr=i;

                 else ed=i;

             }

             {

                 printf("%.4lf\n",i);

             }

         }

     }

     return ;

 }