Can you solve this equation?


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10943    Accepted Submission(s): 5046



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!


 



Author


Redow


题目分析：精度比较恶心的二分，只要不超时，精度尽量开小就对了，涨姿势了
  

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define eps 1e-12
#define EPS 1e-9

using namespace std;

double f ( double x )
{
    return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6.0;
}

bool check ( double x , double n )
{
    if ( f(x)+eps < n ) return true;
}

bool equal ( double a , double b )
{
    if ( fabs ( a -b ) < eps ) return true;
    else return false;
}

double search ( double n )
{
    double left = 0.0 , right = 100.0 , mid;
    while ( !equal ( left , right ))
    {
        mid = (left + right) / 2.0;
        if ( check ( mid , n ) ) left = mid;
        else right = mid;
    }
    return mid;
}

int t;
double n;

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
       scanf ( "%lf" , &n );
       if ( n < f(0) || n > f(100 ) )
       {
           puts ( "No solution!" );
           continue;
       }
       double ans = search ( n );
       printf ("%.4lf\n" , ans );
    }
}