【HD 2199】Can you solve this equation?

Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16574 Accepted Submission(s): 7358
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
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簡単な二分問題で,Yを与え,方程式を与え,forの代わりに二分を用いてxに近づき,最後にxの値を見つけた.最初の2点問題は,記念に残しておく.
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コード:

#include<cstdio>
#include<algorithm>
#include<cmath>

using namespace std;

//8* X^4+ 7* X^3+ 2* X^2+ 3 * X +6= Y
double st( double x)
{
    return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}

int main()
{
    int t,k;
    double l,r;
    double Y,mid;
    scanf("%d",&t);
    while( t-- )
    {
        l=0;r=100,mid=0;
        scanf("%lf",&Y);
        if( Y<st(0) || Y>st(100) )
        {
            printf("No solution!
");
            continue;
        }
        else  
        {
            int SCA = 1000;
            while( SCA -- )
            {
                mid=(r+l)/2.0;
                if( st(mid)>Y )
                {
                    r=mid;
                }
                else 
                    l=mid;
            }
            printf("%.4lf
",l);
        }

    }
    return 0;
}