Hdu 2199--Can you solve this equation? 简单二分

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12645    Accepted Submission(s): 5641

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
   
   
   
   

最简单的二分，但是因为没有注意0<Y<6的情况，WA了10+次，苦逼的人，醉了

#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
const double INF = 0x7fffffff;
const double eps = 1e-10;
int T;
double X,Y,ans;
double Calc(double x) { return (((8.0*x + 7.0)*x + 2.0)*x + 3.0)*x + 6.0; }
void Search()
{
    double MVal,l,r,mid;
    l = 0;r = 101;
    while(r-l >= eps)
    {
        mid = (r+l)/2.0;
        MVal = Calc(mid);
        if(MVal > Y)
        {
            r = mid;
        }
        else if(MVal < Y)
        {
            l = mid;
        }
        else break;
    }
    printf("%.4lf\n",mid);
}
int main() {
    //freopen("input.in","r",stdin);
    double Max = Calc(100);
    for(scanf("%d",&T);T--;)
    {
        scanf("%lf",&Y);
        if(Y<6 || Y>Max) { printf("No solution!\n"); continue; }
        Search();
    }
    return 0;
}