4.1.1--二分--Can you solve this equation？

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

2100-4

Sample Output

1.6152No solution!

代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;#define exp 0.0000000001int main(){    int t,y;    scanf("%d",&t);    while(t--)    {        scanf("%d",&y);        if(y==-6) { printf("0.0000\n"); continue;}        double v,mid,l=0,r=100;        while((r-l)>exp)        {            mid=(r+l)/2;            v=8*pow(mid,4)+7*pow(mid,3)+2*pow(mid,2)+3*mid+6;            if(fabs(v-(double)y)<exp)                break;            else            {                if(v>y) r=mid;                else l=mid;            }        }        if((mid-0)>exp&&(100-mid)>=exp) printf("%.4lf\n",mid);        else printf("No solution!\n");    }    return 0;}