HDU 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): 2091    Accepted Submission(s): 1053

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!

 

 

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cctype>
#include <cstring>
#include <sstream>
#include <fstream>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <algorithm>

using namespace std;
//Constant Declaration
/*--------------------------*/
//#define LL long long 
#define LL __int64
const int M=1000000;
const int INF=1<<30;
const double EPS = 1e-11;
const double PI = acos(-1.0);
/*--------------------------*/
// some essential funtion
/*----------------------------------*/
void Swap(int &a,int &b){   int t=a;a=b;b=t; }
int Max(int a,int b){   return a>b?a:b;  }
int Min(int a,int b){   return a<b?a:b;  }
int Gcd(int a,int b){   while(b){b ^= a ^=b ^= a %= b;}  return a; }
/*----------------------------------*/
//for (i = 0; i < n; i++)
/*----------------------------------*/
#define F(x) (8*pow(x,4) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6 - y)

double y;
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t, case1 = 0;
    scanf("%d", &t);
    int n, m;
    int i, j;
    //scanf("%d%d", &n, &m);

    while (t--)
    {
        scanf("%lf", &y);
        if (F(0.0) * F(100.0) > 0)
        {
            puts("No solution!");
        }
        else
        {
            double l = 0, r = 100;
            if (F(l) == 0)
            {
                printf("%.4lf", 0);
                goto end;
            }
            if (F(r) == 0)
            {
                printf("%.4lf", 100);
                goto end;
            }

            double min;
            while (r - l > EPS)
            {
                min = (l + r) / 2;
                if (F(min) == 0)
                {
                    goto end;
                }
                if (F(l) * F(min) < 0)
                {
                    r = min;
                }
                else
                {
                    l = min;//写成l = min + 1就错
                }     


            }
end : 
            printf("%.4lf\n", min);




        }
    }

    return 0;
}